Ray Optics and Optical Instruments: Class 12 Physics Chapter 9 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

115 min
36 practice questions
27 MCQs

Is chapter mein aap kya seekhenge

Samjho ki light ko kab rays maana ja sakta hai, aur mirrors aur lenses par Cartesian sign convention lagao
f = R/2 aur mirror equation derive karo, aur magnification ke saath unka use karke images ki position aur nature batao
Snell's law lagao, lateral shift aur apparent depth samjhao, aur total internal reflection ke liye critical angles nikaalo
Samjho ki totally reflecting prisms aur optical fibres kaise kaam karte hain
Spherical surface par refraction ka formula, lens maker's formula aur thin lens formula derive karo, aur akele lens aur contact mein rakhe lenses ki power ka use karo
Prism se hone wala deviation derive karo aur angle of minimum deviation se refractive index nikaalo
Simple microscope, compound microscope aur refracting aur reflecting telescopes ki magnifying power samjhao aur calculate karo

NCERT Physics Class 12, Chapter 9, pp. 221–254 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

9.1 Introduction

Tin ki chhat ke ched se tirchhi aati dhoop, ya dhool bhare kamre mein torch ki beam, hawa mein ek sharp seedhi line kheench deti hai. Shadows ke edges saaf hote hain, aur ek oonchi building Sun ko theek tab chhupa leti hai jab woh tumhari aankh se Sun tak ki seedhi line ko block karti hai. Roz ka experience light ke baare mein do baatein batata hai: woh bahut tez chalti hai, aur seedhi lines mein chalti hai. Yeh chapter doosre idea par mirrors, lenses, prisms aur instruments ke liye poora toolkit banata hai.

Part I ke Chapter 8 mein tumne light ko electromagnetic wave ke roop mein dekha tha. Hamari aankhein us spectrum ke ek patle hisse par respond karti hain, lagbhag $400 n m$ (violet) se lagbhag $750 n m$ (red) tak ki wavelengths, aur us hisse ko hum light (प्रकाश) kehte hain. Vacuum mein yeh $c = 2.99792458 \times 10^{8} m / s$ se chalti hai, jo value ab definition se fixed hai; lagbhag har calculation ke liye $3.00 \times 10^{8} m / s$ kaafi hai. Isse tez koi cheez energy ya information nahi le jaati. Phir bhi sunlight ko Earth tak $1.5 \times 10^{11} m$ tay karne mein lagbhag $500 s$ lagte hain.

Ek wave seedhi lines mein kaise chal sakti hai? Jawab hai size. Kisi opening se guzarti wave tabhi saaf-saaf phailti hai jab opening uski wavelength jitni ho. Visible light ka $λ \approx 5 \times 10^{- 7} m$ hai, jabki window, lens ya coin ek centimetre se ek metre tak chauda hota hai, yaani ek lakh se das lakh guna bada. Aisi cheezon ke paas phailaav itna kam hota hai ki dikhta hi nahi, aur energy seedhi lines mein chalti hai (Fig. 9.1). Openings sach mein bahut chhoti ho jaayein to kya hota hai, yeh Chapter 10 mein hai.

3/3Chalne ki direction mein, wavefronts ke perpendicular kheenchi lines rays hain; unka bundle beam kehlata hai.

Fig. 9.1Jab openings wavelength ke muqable bahut badi hoti hain, to light energy seedhi lines, yaani rays, ke along chalti hai.

Left mein ek chhota lamp circular wavefronts bhejta hai. Beech mein chaudi window wali ek deewar hai, aur right mein doosri deewar. Step 1: wavefronts lamp se phailte hain aur jo hissa window se guzarta hai woh arcs ke roop mein aage badhta hai. Step 2: window ke aage ki light seedhe edges wale wedge ki tarah shade ki gayi hai, jo door wali deewar par sharp boundary wala bright patch banati hai, jis par sharp edge likha hai. Step 3: lamp se window ke through deewar tak teen arrows rays ke roop mein kheenche gaye hain, wavefronts ke perpendicular, aur label kehta hai ki beam rays ka bundle hai.

Definition

Ray

Ek seedhi line jiske along light energy chalti hai, wavefronts ke perpendicular kheenchi jaati hai. Rays ke bundle ko beam kehte hain: parallel, converging ya diverging.

ExampleDhundh bhare kamre mein laser pointer jo patli bright line banata hai, woh ek single ray ki achhi picture hai.

Ray picture se hum is chapter mein teen cheezein padhenge: reflection, refraction aur plane aur curved surfaces par inse banne wali images. Pehle spherical mirrors, phir flat surfaces par refraction, total internal reflection, curved surfaces aur thin lenses, aur prism. Chapter simple microscope, compound microscope aur telescope par khatam hota hai, jo in parts se bane instruments hain.

Quick check

3 mein se 1

Moon, Earth se

3.84 \times 10^{8} m

door hai. Moonlight ko hum tak pahunchne mein kitne seconds lagte hain?

Answer dekho

Answer 1.28 s

t = d / c = 3.84 \times 10^{8} / 3.00 \times 10^{8} = 1.28 s

. Moon par rakhe reflector ko bheja gaya laser pulse lagbhag

2.56 s

baad wapas aata hai.

9.2 Spherical mirrors se light ka reflection

Reflection ke do laws tum pehle se jaante ho. Angle of reflection angle of incidence ke barabar hota hai, dono normal se naape jaate hain; aur incident ray, normal aur reflected ray ek hi plane mein hote hain. In laws mein kahin nahi likha ki surface flat honi chahiye. Curved mirror ke kisi bhi point par, us point ke aas-paas ka chhota sa patch lagbhag flat hota hai, isliye laws har point par us patch ke normal ke saath lagte hain.

Spherical mirror, yaani hollow sphere se kata ek tukda, ke har point par normal dhoondhna aasaan hai: woh us point se guzarne wale radius ke along hota hai, isliye har normal centre of curvature C se guzarta hai. Mirror ki surface ka beech wala point uska pole P hai, distance PC radius of curvature R hai, aur P aur C se guzarne wali line principal axis hai. Jis mirror ki hollow side par silvering ho woh concave hai; jiski ubhri side par silvering ho woh convex hai. Lens ke liye iska beech wala point optical centre hai, aur uski principal axis optical centre aur principal foci se guzarne wali line hai.

Word image mein ek aur shart chhupi hai. Polished mirror regularly reflect karta hai: paas-paas ki rays paas-paas ke aise patches par padti hain jo ek hi taraf face karte hain, isliye ek object point se aayi rays saath rehti hain. Safedi wali deewar us par padne wali zyadatar light reflect karti hai, phir bhi tum usmein apna chehra nahi dekh sakte: uske khurdure dane har direction mein face karte hain, isliye ek point se aayi rays har taraf bikhar jaati hain aur kabhi ek image point par nahi milti.

Try karke dekho

Ek chammach mein do mirrors

Tumhe chahiye

ek polished steel ki chammach (badi serving spoon sabse achha kaam karti hai)
kamre ke us paar ek window ya lamp

Pehle guess karo

Chammach ko haath bhar door pakdo aur uski hollow side mein dekho. Tumhara chehra kaisa dikhega?

ulta aur chhotaseedha aur badaseedha aur same size

Chammach ko haath bhar door pakdo, hollow side apni taraf rakho aur apna chehra dekho.
Chammach ko dheere-dheere ek aankh ki taraf lao jab tak woh kuch centimetre door na rahe. Dhyaan do ki image kis pal badalti hai.
Chammach ko palto aur bowl ki peeth dekho, paas se aur door se.

Table 9.1Tumne kya dekha

Kahan dekh rahe ho	Seedha ya ulta?	Bada ya chhota?
hollow side, haath bhar door
hollow side, bahut paas
bowl ki peeth

Tumhari entries apne aap save hoti hain.

9.2.1 Sign convention

Mirror formula ko mirror ke saamne ke objects aur peeche ki images, concave aur convex mirrors, real aur virtual images, sab cover karne hain. Har case ke liye alag formula ki jagah hum har distance ko ek sign dete hain aur poore samay ek hi formula use karte hain. Yeh book Cartesian sign convention use karti hai (Fig. 9.2):

Origin. Mirror ya lens ke beech se naapo, yaani uske pole P se (lens ke liye uske optical centre O se), aur principal axis ko $x$ -axis maano.
Distances. Origin se incident light ke saath chal kar pahunche point ki distance positive hoti hai; light ke against chal kar pahunche point ki distance negative hoti hai.
Heights. Axis ke upar wale points ki heights positive aur neeche wale points ki heights negative hoti hain.

4/4Upar ki heights positive aur neeche ki heights negative hain, isliye inverted image ki height negative hoti hai.

Fig. 9.2Pole se naapo; incident light ki direction positive hai.NCERT Fig. 9.2 · p. 222

Light left se ek convex mirror ki taraf aati hai jiska pole P drawing ke beech mein hai. Step 1: P se right ki taraf, yaani incident light ke along, distances ko arrows positive aur left ki distances ko negative mark karte hain. Step 2: axis par left mein ek object khada hai; P se uski distance left point karte arrow se dikhayi gayi hai aur us par u less than zero likha hai. Step 3: mirror ke peeche ek chhoti seedhi dashed image dikhti hai, aur focus F usse bhi peeche hai; dono distances right point karti hain aur un par v greater than zero aur f greater than zero likha hai. Step 4: ek key dikhati hai ki upar ki heights positive aur neeche ki negative hain, aur object height par h greater than zero likha hai.

Definition

Cartesian sign convention

Distances pole (mirror) ya optical centre (lens) se naapi jaati hain: incident light ki direction mein positive, uske against negative. Principal axis ke upar heights positive, neeche negative.

ExampleLeft se aati light ke saath, mirror ke 30 cm saamne rakhe real object ka $u = - 30 c m$ hai; 15 cm focal length wale concave mirror ka $f = - 15 c m$ hai.

Hum incident light ko hamesha left se right chalta hua banate hain. Tab real object left mein hota hai aur uska $u$ negative hota hai; concave mirror ka focus uske saamne hota hai, isliye uska $f$ negative hai; convex mirror ka focus uske peeche hota hai, isliye uska $f$ positive hai. Image real hai ya nahi, yeh dekh kar signs tay mat karo. Numbers ko unke signs ke saath formula mein daalo, aur answer ka sign batayega ki image kahan hai.

9.2.2 Spherical mirrors ki focal length

Concave mirror par principal axis ke parallel ek patli beam bhejo. Agar rays paraxial hain (mirror ko pole ke paas hit karti hain aur axis se bas thoda sa jhuki hain), to sab axis par ek hi point F se reflect hokar guzarti hain: principal focus. Convex mirror aisi beam ko phaila deta hai, aur reflected rays uske peeche ke ek point F se aati hui lagti hain (Fig. 9.3). Distance PF focal length $f$ hai.

Axis se jhuki parallel beam bhi ek point par aati hai, lekin axis se hat kar. Aise saare points F se guzarne wale, axis ke perpendicular plane mein hote hain, jise focal plane kehte hain. Isiliye door ka scene, jiske points thodi alag directions mein parallel beams bhejte hain, focus ke paas ek flat image banata hai.

1/3Axis ke parallel paraxial rays mirror ke saamne principal focus F se reflect hokar guzarti hain.

Fig. 9.3Concave mirror parallel paraxial beam ko focus par laata hai; convex mirror use ek point se diverge karata hai.NCERT Fig. 9.3 · p. 223

Principal axis figure ke aar-paar jaati hai. Step 1: axis ke parallel chaar rays right mein ek concave mirror par padti hain aur sab point F se reflect hokar guzarti hain, jo pole P aur centre of curvature C ke theek beech mein hai; focal length f, F se P tak mark ki gayi hai. Step 2: wahi rays convex mirror par padti hain aur bahar ki taraf reflect hoti hain; reflected rays ke dashed backward extensions mirror ke peeche F par milte hain. Step 3: ek parallel beam axis se angle par aati hai aur concave mirror se reflect hone ke baad F se guzarti dashed vertical line par axis ke neeche ek point par milti hai, jis par focal plane likha hai.

DerivationFocal length radius ki aadhi kyun hai

Axis ke parallel ek ray lo jo concave mirror par Q par padti hai (Fig. 9.4). Radius CQ, Q par normal hai; angle of incidence ko $θ$ kaho.

Fig. 9.4Normal CQ, C par axis ke saath angle

θ

banata hai, aur reflected ray Q par normal ke saath

θ

banati hai: triangle CFQ ke do angles barabar hain.NCERT Fig. 9.4 · p. 224

Right mein ek concave mirror arc hai aur uska centre of curvature C left mein axis par hai. Axis ke parallel ek ray mirror par Q par padti hai. Dashed normal CQ, jiski length R hai, C par axis ke saath angle theta banata hai. Q par incident ray aur reflected ray dono normal ke saath angle theta banati hain. Reflected ray axis ko F par kaatti hai, jahan woh axis ke saath angle 2 theta banati hai, aur neeche left ki taraf aage jaati hai. Pole P wahan hai jahan axis mirror se milti hai.

Incident ray axis ke parallel hai, isliye normal CQ axis se usi angle $θ$ par milta hai (alternate angles): $∠ F C Q = θ$ . Law of reflection ke kaaran reflected ray QF normal ki doosri taraf $θ$ par hoti hai: $∠ C Q F = θ$ . Isliye triangle CFQ ke do angles barabar hain, to woh isosceles hai:

C F = F Q

(F par exterior angle $2 θ$ hai, isiliye reflected ray axis ko angle of incidence ke double angle par kaatti hai.) Ab paraxial condition lagao. Jab Q pole ke paas ho, to tirchhi length FQ lagbhag FP ke barabar hoti hai, isliye $C F \approx F P$ : focus C aur P ke theek beech mein hota hai. Kyunki $C P = R$ aur $F P = f$ ,

spherical mirror ki focal length

f = \frac{R}{2}

(9.3)

Convex mirror ke liye yahi argument F ko P aur C ke beech, mirror ke peeche rakhta hai. Signs ke saath, concave mirror ke $R$ aur $f$ dono negative hote hain aur convex mirror ke dono positive, aur $f = R / 2$ dono cases mein sahi hai. Yeh argument approximation ki keemat bhi dikhata hai. Axis se door ki ray ke liye FQ, FP se saaf lamba hota hai, isliye F mirror ki taraf khisak jaata hai aur chauda spherical mirror dhundhla focus deta hai. Paraxial rays isse bachti hain; torch ke reflectors parabolic shape use karke isse bachte hain.

9.2.3 Mirror equation

Object ke kisi point se light har direction mein phailti hai. Agar reflection ke baad woh saari rays sach mein ek point se guzarti hain, to woh point object point ki real image (वास्तविक प्रतिबिंब) hai; agar woh peeche badhane par sirf ek point se diverge hoti lagti hain, to woh virtual image (आभासी प्रतिबिंब) hai. Image reflection ya refraction se bani object ki point-to-point copy hai.

Image point dhoondhne ke liye humein sirf do rays chahiye, kyunki do lines apna crossing point fix kar deti hain. Concave mirror ki chaar rays banana aasaan hai (Fig. 9.5):

axis ke parallel ray F se reflect hokar guzarti hai (convex mirror ke liye, aise reflect hoti hai jaise F se aa rahi ho);
F se guzarti ray (ya convex mirror ke F ki taraf jaati ray) axis ke parallel reflect hoti hai;
C se guzarti ray (ya convex mirror ke C ki taraf jaati ray) mirror par normal ke along padti hai aur seedhi wapas aati hai;
pole par padti ray axis ki doosri taraf barabar angle par reflect hoti hai.

5/5P par padti ray symmetrically reflect hoti hai aur image se bhi guzarti hai: yahan image object se double size ki hai.

Fig. 9.5Chaar aasaan rays mein se koi bhi do image fix kar deti hain; baaki bhi usi point se guzarti hain.NCERT Fig. 9.5 · p. 224

Right mein ek concave mirror hai, axis par uska pole P, focus F aur centre of curvature C hai, aur F aur C ke beech ek seedha object arrow hai. Step 1 sirf yeh set-up dikhata hai. Step 2: object ke top se ek blue ray axis ke parallel chalti hai, reflect hoti hai aur F se guzarti hai. Step 3: object ke top se ek orange ray F se guzarti hai, mirror se takrati hai aur axis ke parallel wapas aati hai; jahan yeh blue ray ko kaatti hai, C se aage, wahan ek ulta image arrow dikhta hai. Step 4: C se object ke top tak ki line ke along ek green ray mirror par padti hai aur C se hokar seedhi usi image point par wapas aati hai. Step 5: ek purple ray pole par padti hai aur axis ke neeche barabar angle par reflect hoti hai, yeh bhi image ke top se guzarti hai. Note mein likha hai real, inverted, double lambi.

Diagram mein bas kuch rays dikhayi gayi hain, lekin object point anginat rays bhejta hai. Mirror tak pahunchne wali har ray usi image point se guzarti hai, aur isi se image sharp banti hai. Ab hum drawing ko equation mein badalte hain.

DerivationMirror equation

Pole ko origin par rakho, axis ko $x$ -axis maano aur light $+ x$ direction mein chal rahi hai. Object ka tip $(u, h)$ par, image ka tip $(v, h^{'})$ par aur centre of curvature $(R, 0)$ par ho, sab apne signs ke saath. Hum Fig. 9.5 ki chaar rays mein se do use karte hain.

Pole par jaati ray. P par normal khud axis hai, isliye reflection ray ki horizontal direction ulat deta hai lekin vertical nahi: reflected ray usi dhalaan $h / ∣ u ∣$ se neeche aati rehti hai. Incident ray $x = u$ par height $h$ se axis tak pahunchti hai; reflected ray ki $x = v$ par height $h^{'} = - h v / u$ hai. Isliye

\frac{h^{'}}{h} = - \frac{v}{u}

C se guzarti ray. Yeh radius ke along chalti hai, mirror par normally padti hai aur apna raasta wapas tay karti hai, isliye object ka tip, C aur image ka tip $(R, 0)$ se guzarne wali ek hi seedhi line par hote hain. C se guzarti seedhi line ke along heights C se distance ke proportion mein badhti hain:

\frac{h^{'}}{h} = \frac{v - R}{u - R}

Dono ek hi image ko describe karte hain, isliye $- \frac{v}{u} = \frac{v - R}{u - R}$ . Cross-multiply karne par $- u v + v R = u v - u R$ , yaani $u R + v R = 2 u v$ . $u v R$ se divide karo aur $R = 2 f$ use karo:

mirror equation

\frac{1}{v} + \frac{1}{u} = \frac{2}{R} = \frac{1}{f}

(9.7)

jahan

$u$: object distance
$v$: image distance
$f$: focal length, R/2

Kisi bhi step mein yeh jaanne ki zaroorat nahi padi ki image real thi ya virtual, ya mirror kis tarah ka tha: yeh information signs mein hai. Isliye (9.7) concave aur convex mirrors, aur real aur virtual images, sab ke liye sahi hai, bas har quantity apne sign ke saath daalni hai.

Image object ke muqable kitni lambi hai, yeh linear magnification se naapa jaata hai:

magnification

m = \frac{h^{'}}{h}

(9.8)

mirror ka magnification

m = - \frac{v}{u}

(9.9)

Real image ke liye $u$ aur $v$ dono negative hote hain, isliye $m$ negative hota hai: single mirror se bani real images inverted hoti hain. Virtual image ke liye $v$ positive hai, $m$ positive hai aur image erect hai. $∣ m ∣$ ka size batata hai ki image enlarged ( $∣ m ∣ > 1$ ) hai ya diminished. Fig. 9.6 roz ki do virtual images dikhata hai: F ke andar object wala concave mirror (shaving ya make-up mirror) aur convex mirror (auto-rickshaw ka mirror).

Fig. 9.6Dono images mirror ke peeche aur seedhi hain; concave wali badi hai, convex wali chhoti.NCERT Fig. 9.6 · p. 226

Do panels. Upar: ek concave mirror jiske focus F aur pole P ke beech object hai. Axis ke parallel ray F se reflect hokar guzarti hai, aur F ki direction se aati ray parallel reflect hoti hai; unke dashed backward extensions mirror ke peeche ek badi, seedhi, dashed image ke top par milte hain, jis par virtual likha hai. Neeche: ek convex mirror jiske kaafi aage object hai. Axis ke parallel ray aise reflect hoti hai jaise mirror ke peeche ke focus F se aa rahi ho, aur centre of curvature ki taraf bheji ray seedhi wapas aati hai; unke dashed extensions mirror ke peeche, pole aur F ke beech, ek chhoti seedhi dashed image ke top par milte hain.

InteractiveMirror bench

Ek interactive ray diagram. Do buttons se concave ya convex mirror chuna jaata hai. Sliders object distance ko 4 se 60 centimetre aur focal length ko 5 se 25 centimetre tak set karte hain. Drawing mein mirror, uska focus aur centre of curvature, 2 centimetre lamba object, parallel ray, pole par jaati ray aur radius ke along ray dikhti hai, virtual image hone par dashed extensions ke saath, aur khud image bhi. Readouts signed object distance, image distance, focal length, magnification aur image height dete hain, aur ek status line image ko real ya virtual, inverted ya erect, enlarged ya diminished batati hai.

Fig. 9.7Concave ya convex mirror chuno, phir object ko move karo aur focal length badlo. u, v, f aur m ki signed values padho.

Mirror bench try karo: concave mirror se shuru karo aur object ko door se mirror ki taraf khiskao. Real image badi hoti hai aur bahar ki taraf jaati hai, object ke F tak pahunchte hi infinity ki taraf bhaag jaati hai, aur mirror ke peeche ek badi virtual image ke roop mein wapas dikhti hai. Convex mirror kabhi kuch dramatic nahi karta: uski image hamesha virtual, seedhi aur chhoti hoti hai.

Example 9.1NCERT Example 9.1 · p. 226

Shaving mirror par paper ki disc

Ek concave shaving mirror deewar par lamp ki sharp real image banata hai. Mirror ke beech mein paper ki ek circular disc chipka di jaati hai jiska radius mirror ke radius ka aadha hai. Deewar par image ka kya hota hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9.2NCERT Example 9.2 · p. 226

Axis ke along padi pencil

Ek $5.0 c m$ pencil concave mirror ( $f = 10 c m$ ) ki axis ke along padi hai, uska tip mirror se $25 c m$ aur blunt end $30 c m$ door hai. Uski image ki length nikaalo aur samjhao ki image distorted kyun hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 9.825 cm aur 30 cm ke beech padi 5 cm ki pencil ki image 15 cm aur 16.7 cm ke beech 1.7 cm ka patla tukda banti hai.NCERT Fig. 9.7 · p. 227

Right mein ek concave mirror hai, jisse F 10 centimetre aur C 20 centimetre door hai. 5 centimetre lambi ek pencil axis par mirror se 25 aur 30 centimetre ke beech padi hai, uska nukeela tip mirror ki taraf hai. Uski image C aur F ke beech hai, sirf 1.7 centimetre lambi, patli, aur uska tip mirror se door point karta hai.

Example 9.3NCERT Example 9.3 · p. 227

Concave mirror ke saamne do positions

Ek concave mirror ka radius of curvature $24 c m$ hai. Ek object uske (a) $18 c m$ aur (b) $8.0 c m$ saamne rakha jaata hai. Har baar image ki position nikaalo, batao ki woh real hai ya virtual, aur $m$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9.4NCERT Example 9.4 · p. 227

Rear-view mirror mein image kitni tez chalti hai?

Ek auto-rickshaw khada hai aur uska convex rear-view mirror ( $R = 1.6 m$ ) ek scooter ki taraf hai jo steady $10 m / s$ se paas aa raha hai. Jab scooter mirror se (a) $50 m$ , (b) $10 m$ aur (c) $2.0 m$ door ho, to scooter ki image kitni tez chalti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.3 Refraction

Nimbu-pani ke glass mein straw daalo to woh surface par tooti hui dikhti hai. Jab light do transparent media ki boundary par pahunchti hai, to uska kuch hissa reflect hota hai aur baaki doosre medium mein chala jaata hai. Agar woh tirchhi aaye ( $0^{\circ} < i < 90^{\circ}$ ), to transmitted light boundary par direction badal leti hai. Direction ke is badlaav ko refraction (अपवर्तन) kehte hain, aur iske rules kisi ke samjhane se bahut pehle measurement se mil gaye the (Fig. 9.9).

LawSnell's laws of refraction

(i) Refracted ray us plane mein rehti hai jo incident ray aur takkar wale point par normal se banta hai. (ii) Media ke ek diye hue pair aur light ke ek diye hue colour ke liye, $\sin i / \sin r$ har angle of incidence par same value leta hai.

Snell's law

\frac{\sin i}{\sin r} = n_{21}

(9.10)

jahan

$i$: angle of incidence, normal se
$r$: angle of refraction, normal se
$n_{21}$: medium 1 ke relative medium 2 ka refractive index

Constant $n_{21}$ medium 1 ke relative medium 2 ka refractive index (अपवर्तनांक) hai. Yeh media ke pair ka hota hai aur colour (wavelength) ke saath thoda badalta hai, lekin angle par depend nahi karta. Agar $n_{21} > 1$ , to $r < i$ : ray normal ki taraf mudti hai, aur medium 2 ko optically denser kehte hain. Agar $n_{21} < 1$ , to optically rarer medium mein jaate hue ray normal se door mudti hai.

2/3Zyadatar light normal ki taraf mud kar water mein jaati hai ( $r < i$ ); thodi si reflect hoti hai.

Fig. 9.9Denser water mein ray normal ki taraf mudti hai; ulta chalne par woh usi raaste se normal se door mudti hai.NCERT Fig. 9.8 · p. 228

Upar air, neeche water, beech mein horizontal boundary aur ek dashed vertical normal. Step 1: hawa mein ek ray normal se angle i par boundary par pahunchti hai. Step 2: water mein ray chhote angle r par aage badhti hai, normal ki taraf mudi hui, aur ek kamzor dashed ray hawa mein wapas reflect hoti hai. Step 3: arrows ulat diye gaye hain: water se angle i par upar aati ray bade angle r par air mein refract hoti hai, usi raaste par normal se door mudte hue.

Refraction isliye hota hai kyunki light alag media mein alag speeds se chalti hai. Kisi medium ka absolute refractive index $n = c / v$ hai, jahan $v$ usmein light ki speed hai; vacuum ke liye $n = 1$ aur air ke liye $n = 1.0003$ , jise hum 1 lete hain. Absolute indices ke terms mein $n_{21} = n_{2} / n_{1} = v_{1} / v_{2}$ , aur Snell's law symmetric form $n_{1} \sin i = n_{2} \sin r$ le leta hai. Water ( $n = 1.33$ ) light ko $2.26 \times 10^{8} m / s$ tak dheema karta hai; diamond ( $n = 2.42$ ) use $1.24 \times 10^{8} m / s$ tak dheema karta hai.

Kyunki light ka raasta hamesha ulta chalaya ja sakta hai (Fig. 9.9, step 3), medium 2 ke relative medium 1 ka index bas reciprocal hai:

n_{12} = \frac{1}{n_{21}}

(9.11)

Relative indices ek chain mein bhi jud jaate hain. Kyunki $n_{32} = n_{3} / n_{2} = (n_{3} / n_{1}) (n_{1} / n_{2})$ , humein milta hai $n_{32} = n_{31} \times n_{12}$ . Jaise, water ke relative glass ka index $n_{glass, water} = 1.50 / 1.33 = 1.13$ hai: water se glass mein jaati ray bas thoda sa mudti hai.

Parallel-sided slab. Glass block ko paar karti ray do baar refract hoti hai (Fig. 9.10). Dono faces parallel hain, isliye normals parallel hain aur ray doosre face se usi angle $r$ par milti hai jis par usne pehla face chhoda tha. Exit par Snell's law, $n \sin r = \sin e$ , phir $e = i$ deta hai. Ray usi direction mein nikalti hai jismein aayi thi: koi deviation nahi, bas side mein ek lateral shift $d$ . Figure ki geometry se, $t$ motaai wale slab ke liye $d = t \sin (i - r) / \cos r$ . $n = 1.50$ aur $i = 50^{\circ}$ ke saath $6.0 c m$ ka slab $r = {30.7}^{\circ}$ aur $d = 2.3 c m$ deta hai.

Fig. 9.10Ray slab se apni original direction mein nikalti hai, side mein

d

khiski hui.NCERT Fig. 9.9 · p. 229

t motaai ka ek rectangular glass slab, upar aur neeche air. Ek ray top face mein angle i par ghusti hai, glass mein chhote angle r par chalti hai, aur bottom face se i ke barabar angle e par nikalti hai. Ek dashed line original ray ko glass ke through seedha aage badhati hai; emergent ray uske parallel chalti hai, perpendicular distance d se khiski hui, jo double-headed arrow se mark hai.

Apparent depth. Paani bhari balti ke tal par pada coin surface ke jitna asal mein hai usse zyada paas dikhta hai. Coin se aati rays water se nikalte hue normal se door mudti hain, isliye jab tumhari aankh unhe seedhi lines mein peeche trace karti hai to woh upar ke ek point se aati lagti hain (Fig. 9.11).

3/3Seedhi lines mein peeche trace karne par (dashed) woh coin ke upar milti hain: aankh use apparent depth par dekhti hai.

Fig. 9.11Upar se dekhne par coin utha hua dikhta hai: apparent depth = real depth / n.NCERT Fig. 9.10 · p. 229

Side se dikhta paani ka tumbler, tal par ek coin aur upar ek aankh. Step 1: do rays coin se vertical ke saath chhote angles par nikalti hain aur surface tak pahunchti hain. Step 2: surface par air mein jaate hue woh normal se door mudti hain aur aankh tak jaati hain. Step 3: dashed lines air wali rays ko peeche water mein badhati hain, jahan woh coin ke upar ek point par milti hain, jis par image likha hai. Left mein ek double arrow surface se coin tak real depth dikhata hai; right mein ek chhota arrow image tak apparent depth dikhata hai.

DerivationLagbhag normal se dekhne par apparent depth

Ek ray real depth $h_{r}$ par pade coin se vertical ke saath chhote angle $θ$ par nikalti hai aur coin ke theek upar wale point se horizontal distance $x = h_{r} \tan θ$ par surface tak pahunchti hai. Air mein woh angle $θ^{'}$ par chalti hai, jahan $\sin θ^{'} = n \sin θ$ . Peeche trace karne par woh coin se guzarti vertical ko depth $h_{a}$ par kaatti hai, jahan $x = h_{a} \tan θ^{'}$ . $x$ ke dono expressions barabar karo aur chhote angles ke liye $\tan \approx \sin$ use karo:

h_{a} = h_{r} \frac{\tan θ}{\tan θ^{'}} \approx h_{r} \frac{\sin θ}{\sin θ^{'}} = \frac{h_{r}}{n}

Result $θ$ par depend nahi karta, isliye lagbhag vertical saari rays ek hi point se aati lagti hain: saaf image. $2.0 m$ gehra swimming pool kinare se sirf $1.5 m$ gehra dikhta hai, jo tairna na jaanne walon ke liye ek jaana-maana khatra hai. Tirchha dekhne par rays ek point par agree nahi karti aur image aur bhi upar uthti hai.

Try karke dekho

Coin ko wapas dikhao

Tumhe chahiye

ek opaque steel ka tumbler ya mug
ek coin
paani ka ek jug

Pehle guess karo

Tum apni aankh aise rakhte ho ki khaali tumbler ka kinara coin ko bas chhupa de. Agar koi coin ko hilaye bina dheere-dheere paani daale, to kya hoga?

Coin chhupa rehta haiCoin dikhne lagta haiCoin aur neeche dhansta lagta hai

Coin ko table par rakhe khaali tumbler ke beech mein rakho.
Apna sir peeche le jao jab tak kinara coin ko bas chhupa na de. Sir ko hilne mat do (thodi ko haathon par tika lo).
Kisi dost se kaho ki coin ko chhede bina dheere-dheere paani daale, aur dekho.

Quick check

4 mein se 1

Air mein ek ray glass (

n = 1.50

) par normal se

40^{\circ}

par padti hai. Angle of refraction degrees mein nikaalo.

Answer dekho

Answer 25.4 °

\sin r = \sin 40^{\circ} / 1.50 = 0.6428 / 1.50 = 0.4285

, isliye

r = {25.4}^{\circ}

. Ray normal ki taraf mudti hai.

9.4 Total internal reflection

Ab light ko ulti taraf bhejo, denser medium se rarer medium mein, jaise water se air mein. Surface par kuch light wapas water mein reflect hoti hai, jise internal reflection kehte hain, aur baaki normal se door mud kar bahar refract hoti hai, taaki $r > i$ . Jaise-jaise $i$ badhta hai, $r$ usse tez badhta hai, aur ek khaas angle of incidence par refracted ray $r = 90^{\circ}$ ke saath surface ke saath-saath sarakti hai. Woh angle critical angle $i_{c}$ (क्रांतिक कोण) hai (Fig. 9.12).

$i_{c}$ se aage, Snell's law ko $\sin r > 1$ chahiye hoga, jo impossible hai: koi refracted ray hoti hi nahi, aur saari light wapas denser medium mein reflect ho jaati hai. Yeh total internal reflection (पूर्ण आंतरिक परावर्तन) hai. Sabse achha silvered mirror bhi us par padne wali light ka kuch percent absorb kar leta hai; total internal reflection mein lagbhag kuch bhi nahi khota, isiliye yeh itna useful hai.

4/4 $62^{\circ}$ par kuch bhi bahar nahi aata: light totally internally reflect ho jaati hai.

Fig. 9.12Angle of incidence ke critical angle se aage jaate hi refraction ruk jaata hai aur saari light reflect hoti hai.NCERT Fig. 9.11 · p. 230

Ek lamp paani ke neeche hai aur upar air hai. Har step lamp se surface tak ek ray dikhata hai, aur jahan woh pahunchti hai wahan normal bana hai; pichhli rays halki grey lines ke roop mein rehti hain. Step 1: 25 degrees par ray air mein 34 degrees par refract hoti hai aur ek kamzor dashed ray reflect hoti hai. Step 2: 40 degrees par refracted ray normal se aur door, 59 degrees tak mudti hai. Step 3: critical angle par, lagbhag 48.8 degrees, refracted ray surface ke saath-saath chalti hai. Step 4: 62 degrees par koi refracted ray nahi hai aur poori ray wapas neeche water mein reflect hoti hai.

Medium 1 (denser) se medium 2 (rarer) mein jaati light ke liye Snell's law mein $r = 90^{\circ}$ rakho, $n_{1} \sin i_{c} = n_{2}$ :

critical angle

\sin i_{c} = \frac{n_{2}}{n_{1}} = n_{21}

(9.12)

Air mein rakhe index $n$ wale material ke liye yeh $\sin i_{c} = 1 / n$ hai, isliye critical angle naapne se refractive index mil jaata hai: $n = 1 / \sin i_{c}$ . Index jitna bada, critical angle utna chhota (Table 9.2).

Table 9.2Kuch transparent media se air mein nikalti light ke critical angles (

\sin i_{c} = 1 / n

)NCERT Table 9.1 · p. 230

Medium	n	Critical angle
Ice	1.31	49.8°
Water	1.33	48.8°
Acrylic plastic	1.49	42.2°
Crown glass	1.52	41.1°
Dense flint glass	1.62	38.1°
Diamond	2.42	24.4°

InteractiveRefraction lab

Flat boundary se milti ray ka ek interactive diagram. Chhe buttons media ka pair chunte hain: air se water, air se glass, water se air, glass se air, glass se water aur diamond se air. Ek slider angle of incidence ko 0 se 89 degrees tak set karta hai. Drawing incident ray, reflected ray, jiski motaai reflect hone wali light ke fraction ke saath badhti hai, aur refracted ray (jab ho) dikhati hai. Readouts dono refractive indices, angles of incidence aur refraction, products n sin i aur n sin r, critical angle aur reflected percentage dete hain; ek status line batati hai ki ray normal ki taraf mudti hai, usse door mudti hai, ya totally internally reflect hoti hai.

Fig. 9.13Do media chuno aur angle of incidence badlo. Refracted ray, reflect hone wali light ka hissa aur, rarer medium mein jaate hue, critical angle dekho.

DemonstrationNCERT Total internal reflection ka ek demonstration · p. 230

Light ko raasta dikhati paani ki dhaar

Tumhe chahiye

dhakkan wali ek saaf plastic bottle (1 litre)
ek laser pointer (red, low power)
doodh ki kuch boondein
ched karne ke liye ek keel
ek balti ya sink
ek andhera kamra

Pehle guess karo

Bottle ke tal ke paas ek chhote ched se paani ek curved arc mein nikalta hai. Laser bottle ke through seedha ched par aim kiya gaya hai. Red light kahan jaayegi?

seedhi aage, curved dhaar ko turant chhod karsirf wapas bottle meincurved dhaar ke andar-andar, jo chamakti hai

Kisi bade se kaho ki garam keel se bottle ke tal se lagbhag $5 c m$ upar lagbhag $4 m m$ chauda ched kar de.
Bottle mein paani bharo, doodh ki teen-chaar boondein daalo aur hilao. Halka dhundhlapan light ka raasta dikhne laayak bana deta hai.
Andhere kamre mein laser ko bottle ki side se upar ki taraf water surface par chamkao. Do spots dhoondho: ek beam surface se bahar nikalti hai, doosri wapas neeche reflect hoti hai.
Laser ko jhukao taaki woh surface se zyada se zyada tirchhi mile. Note karo ki surface ke upar wali beam kab gayab hoti hai.
Ab bottle ko sink ke upar pakdo, dhakkan kholo taaki dhaar behe, aur laser ko ulti taraf se bottle ke through horizontally ched par aim karo.
Dhaar ko dekho, aur us spot ko jahan woh sink mein girti hai.

Table 9.3Observations

Tumne kya kiya	Tumne kya dekha
Beam surface se steeply milti hai
Beam surface se bahut tirchhi milti hai
Beam water jet mein aim ki

Tumhari entries apne aap save hoti hain.

Steep beam bant jaati hai: zyadatar water se bahar nikalti hai, thodi reflect hoti hai. Ek khaas jhukaav (critical angle, normal se lagbhag $49^{\circ}$ ) ke baad upar wali beam gayab ho jaati hai aur surface ek perfect mirror ki tarah kaam karti hai. Jet mein light har baar curved water surface se glancing angle par milti hai, isliye baar-baar totally reflect hoti hai; jet apni poori length mein chamakta hai aur light wahin girti hai jahan paani girta hai. Optical fibre bilkul isi tarah light ko modon ke paar le jaata hai.

9.4.1 Nature mein total internal reflection aur iske technological applications

(i) Prisms jo perfect mirrors ki tarah kaam karte hain. Jis glass prism ke angles $45^{\circ}$ , $45^{\circ}$ aur $90^{\circ}$ hain, woh bina silvering ke light ko mod sakta hai. Ek face mein normally ghusi light lambe face se $45^{\circ}$ par milti hai, jo glass ke critical angle ( $41$ – $42^{\circ}$ , Table 9.2) se zyada hai, aur totally reflect ho jaati hai. Prism ko kaise ghumaya gaya hai, us hisaab se woh beam ko $90^{\circ}$ mod deta hai, $180^{\circ}$ par wapas bhejta hai, ya image ka size badle bina use ulta kar deta hai (Fig. 9.14). Periscopes aur binoculars aise prisms use karte hain; unke reflections dashakon tak bright rehte hain kyunki unpar koi coating nahi jo kharab ho.

2/3Do total reflections light ko usi raaste wapas bhejte hain jis se woh aayi, jaise bicycle reflector ya binoculars ke prism mein.

Fig. 9.1445° par total internal reflection prism ko loss-free mirror bana deta hai.NCERT Fig. 9.13 · p. 231

Step 1: do rays right-angled prism ke vertical face mein ghusti hain, dhalaan wale lambe face se 45 degrees par milti hain, totally reflect hoti hain aur bottom face se neeche nikalti hain, 90 degrees mud kar. Step 2: prism ko aise ghumaya gaya hai ki lamba face light ki taraf ho; do rays usmein ghusti hain, baari-baari se har chhote face par totally reflect hoti hain aur apni original direction ke parallel wapas aati hain, unka order badal jaata hai. Step 3: prism apne lambe face par leta hai; rays A aur B ek ke upar ek left dhalaan wale face mein ghusti hain, refract hoti hain, base par totally reflect hoti hain aur right dhalaan wale face se nikalti hain, ab B, A ke upar hai.

(ii) Optical fibres. Optical fibre (प्रकाशिक तंतु) baal se bhi patla glass ya plastic ka dhaaga hai. Iske core ka refractive index aas-paas ki cladding se thoda zyada hota hai. Ek end mein sahi angle par ghusi light core–cladding boundary se critical angle se zyada angle par milti hai, totally reflect hoti hai, doosri side jaati hai, aur phir reflect hoti hai, har metre mein hazaaron baar (Fig. 9.15). Fibre ko halka sa moda bhi ja sakta hai: light mod ke saath-saath chalti hai, isiliye ise aksar light pipe kehte hain.

Fig. 9.15Har baar ray core–cladding boundary se milti hai to totally reflect hoti hai, isliye light core mein hi rehti hai.NCERT Fig. 9.14 · p. 232

Cross-section mein kheencha gaya ek lamba seedha fibre: beech mein zyada refractive index wala core, jo kam refractive index wali cladding ki do layers ke beech hai. Ek ray left end se ghusti hai, core paar karti hai, core-cladding boundary se glancing angle par milti hai, totally reflect hoti hai, aur core mein zig-zag chalti rehti hai jab tak right end se nikal na jaaye.

Fibres telephone calls, television aur internet data ko light ki flashes ke roop mein le jaate hain, oceans ke neeche bhi aur gharon tak bhi. Fibres ke bundles light ko body ke andar le jaate hain aur image bahar laate hain: aise endoscope se doctors bina bade cuts ke pet, saans ki nalikaon ya chot khaaye ghutne ke andar dekh sakte hain. Melon mein bikne wale guchhedaar lamps, jinke plastic ke dhaage tips par chamakte hain, decorative optical fibres hain jinhe base ke bulb se roshni milti hai.

Asli challenge purity hai. Aam window glass saaf dikhta hai, lekin kuch metre mota block gehra hara hoga. Fibres ke liye silica itni saavdhaani se pure ki jaati hai ki poore ek kilometre ke baad bhi $95 %$ se zyada light bachi rehti hai.

Example

Light fibre mein kitni steeply ghus sakti hai?

Ek fibre ke core ka index $1.50$ aur cladding ka index $1.46$ hai. (a) Core–cladding boundary par critical angle kya hai? (b) Core ke andar ki ray fibre ki axis ke saath sabse bada kitna angle bana sakti hai aur phir bhi guide ho? (c) Air se flat end mein ghus kar guide hone wali light axis ke saath sabse bada kitna angle bana sakti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.5 Spherical surfaces par aur lenses se refraction

Patte par paani ki boond, glass ki marble aur tumhari aankh ka lens, sab curved surfaces par light refract karte hain. Curved mirrors ki tarah, hum spherical surface ke chhote patch ko flat maante hain, wahan Snell's law lagate hain, aur yaad rakhte hain ki kisi bhi point par normal radius ke along, centre of curvature se hokar jaata hai. Pehle hum dekhte hain ki ek single spherical surface kya karti hai. Phir thin lens bas aisi do surfaces hain, paas-paas, jinmein se kam se kam ek curved hai: single-surface result ko do baar lagane se lens maker's formula aur lens formula milte hain.

9.5.1 Spherical surface par refraction

Medium 1 (index $n_{1}$ , left mein) medium 2 (index $n_{2}$ , right mein) se ek convex spherical boundary par milta hai jiska centre C aur radius $R$ hai. Ek point object O pehle medium mein axis par hai, aur P woh point hai jahan axis surface se milti hai. Humein woh point I chahiye jahan refracted rays axis ko kaatti hain. Aperture ko baaki saari distances ke muqable chhota rakho, taaki Fig. 9.16 ka har angle chhota ho.

DerivationEk refracting surface se bani image

Fig. 9.16O, C aur I par angles

α

β

aur

γ

hain; A par

i = α + β

aur

r = β - γ

.NCERT Fig. 9.15 · p. 233

Principal axis left se right object point O, axis aur surface ke milne ke point P, centre of curvature C aur image point I se hokar jaati hai. Surface axis ke upar ke point A se hokar curve hoti hai; uske right ka hissa doosra medium hai. O se A tak ki ray axis ke saath angle alpha banati hai; radius CA, dashed aur bahar tak badhaya hua, A par normal hai aur C par axis ke saath angle beta banata hai. Andar, A se I tak refracted ray axis ke saath angle gamma banati hai. A par angle of incidence i badhayi hui incident ray aur normal ke beech, aur angle of refraction r refracted ray aur normal ke beech naapa gaya hai.

O se aati ray surface se A par mile, jo axis se chhoti height $h$ upar hai, aur OA, CA aur AI axis ke saath jo angles banate hain unhe $α$ , $β$ aur $γ$ kaho. Har angle chhota hai, isliye woh lagbhag $h$ ko P se horizontal distance se divide karne ke barabar hai:

α \approx \frac{h}{O P}, β \approx \frac{h}{P C}, γ \approx \frac{h}{P I}

O, A aur C ko ek triangle maano: angle of incidence $i$ (aane wali ray, aage badhayi hui, aur normal ke beech) A par iska exterior angle hai, isliye $i = α + β$ . Isi tarah triangle ACI mein C par angle $β$ exterior hai, isliye $β = r + γ$ , yaani $r = β - γ$ :

i = \frac{h}{O P} + \frac{h}{P C}, r = \frac{h}{P C} - \frac{h}{P I}

(9.14)

Chhote angles ke liye Snell's law $n_{1} i = n_{2} r$ hai. Substitute karke aur $h$ cancel karke,

\frac{n_{1}}{O P} + \frac{n_{2}}{P I} = \frac{n_{2} - n_{1}}{P C}

(9.15)

Yeh lengths hain. Cartesian convention ke saath, P se incident light ke along naapne par $O P = - u$ , $P I = + v$ aur $P C = + R$ :

spherical surface par refraction

\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}

(9.16)

jahan

$n_{1}$: us medium ka index jisse light aa rahi hai
$n_{2}$: us medium ka index jismein light ja rahi hai
$R$: radius of curvature, positive agar C outgoing side par ho

Signs ke saath, (9.16) convex aur concave surfaces aur real aur virtual images, sab ke liye sahi hai. Flat surface ke liye $R \to \infty$ aur yeh $v = (n_{2} / n_{1}) u$ deta hai, jo Section 9.3 ka apparent-depth result hi hai.

Example 9.5NCERT Example 9.5 · p. 233

Curved glass surface par images

(a) Air mein ek point source ek lambi glass rod ( $n = 1.50$ ) ki convex surface se $40 c m$ door hai, jiska end $10 c m$ radius tak ghisa gaya hai. Image kahan banti hai? (b) Ek solid glass marble ke theek centre mein hawa ka chhota bubble hai. Bahar se marble ko dekhne wale ko woh kahan dikhta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.5.2 Lens se refraction

Index $n_{1}$ wale medium mein rakhe index $n_{2}$ ke thin lens ki do surfaces hain, radii $R_{1}$ (pehli surface jisse light milti hai) aur $R_{2}$ . Trick yeh hai ki pehli surface ko akele kaam karne do, jaise uske aage saari jagah glass bhara ho, aur woh image $I_{1}$ nikaalo jo woh banati. Phir woh image doosri surface ke liye object ka kaam karti hai (Fig. 9.17).

3/3Surface 2 rays ko phir modti hai aur woh final image $I$ par milti hain.

Fig. 9.17Pehli surface

I_{1}

banati; doosri surface

I_{1}

ko object maan kar final image

I

banati hai.NCERT Fig. 9.16 · p. 234

Double convex lens ke left mein ek point object O. Step 1: sirf pehli surface bani hai, aur uske right ki saari jagah glass se bhari hai; O se aayi ray is surface par refract hoti hai aur right mein kaafi door point I1 ki taraf jaati hai. Step 2: poora lens bana hai; glass ke andar ki ray ab bhi I1 ki taraf jaati hai (lens ke aage dashed), isliye I1 doosri surface ke liye virtual object hai. Step 3: doosri surface par ray phir refract hoti hai aur lens ke kaafi paas final image I par axis ko kaatti hai.

DerivationLens maker's formula aur thin lens formula

Pehli surface ( $n_{1}$ se $n_{2}$ mein, radius $R_{1}$ ), object distance $u$ par, image $I_{1}$ $v_{1}$ par. (9.16) se:

\frac{n_{2}}{v_{1}} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R_{1}}

(9.17)

Doosri surface ( $n_{2}$ se wapas $n_{1}$ mein, radius $R_{2}$ ). Iska object $I_{1}$ hai. Thin lens ke liye dono surfaces lagbhag optical centre par hain, isliye object distance phir $v_{1}$ hai; final image $v$ par hai. Media badal kar (9.16) lagao:

\frac{n_{1}}{v} - \frac{n_{2}}{v_{1}} = \frac{n_{1} - n_{2}}{R_{2}}

(9.18)

Dono equations jodne se $v_{1}$ hat jaata hai:

\frac{n_{1}}{v} - \frac{n_{1}}{u} = (n_{2} - n_{1}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

(9.19)

Infinity par rakhe object ( $u \to - \infty$ ) ki image second principal focus par banti hai, $v = f$ . $n_{1}$ se divide karo aur $n_{21} = n_{2} / n_{1}$ likho:

lens maker's formula

\frac{1}{f} = (n_{21} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

(9.21)

Isliye (9.19) ka right-hand side $n_{1}$ se divide karne par $1 / f$ hai, aur

thin lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

(9.23)

Double convex lens ke liye $R_{1} > 0$ aur $R_{2} < 0$ , isliye $f > 0$ : yeh light ko converge karta hai. Double concave lens ke liye signs ulat jaate hain ( $R_{1} < 0$ , $R_{2} > 0$ ) aur $f < 0$ : yeh light ko diverge karta hai. Kisi bhi formula ko yeh jaanne ki zaroorat nahi padi ki hum kis case mein the.

Lens maker's formula (9.21) lens designer use karta hai: ek glass chuno, phir aise radii chuno jo chahi gayi focal length dein. Lens ke do principal foci hote hain, aane wali light ki side par F (first focal point) aur doosri side par F′ (second focal point). Jab thin lens ki dono taraf same medium ho, to dono optical centre O se same distance $∣ f ∣$ par hote hain.

Drawing ke liye thin lens se teen rays aasaan hain (Fig. 9.18):

axis ke parallel ray convex lens ke second focus F′ se guzarti hai, ya concave lens ke first focus F se aati hui diverge hoti hai;
optical centre O se guzarti ray bina mude seedhi aage jaati hai;
convex lens ke F se guzarti ray (ya concave lens ke F′ ki taraf jaati ray) axis ke parallel nikalti hai.

4/5F se guzarti ray parallel nikalti hai; teeno milti hain: 2F′ se aage real, inverted, enlarged image.

Fig. 9.18Teen aasaan rays lens ki image fix karti hain; concave lens ke liye unke backward extensions milte hain.NCERT Fig. 9.17 · p. 235

Step 1: ek convex lens, uska optical centre O, foci F aur F prime aur points 2F aur 2F prime axis par mark hain; 2F aur F ke beech ek seedha object khada hai. Step 2: object ke top se ek blue ray axis ke parallel chalti hai aur lens ke baad F prime se guzarti hai. Step 3: ek green ray O se seedhi guzarti hai. Step 4: F se guzarti ek orange ray axis ke parallel nikalti hai; teeno rays 2F prime se aage milti hain, jahan ek ulti, badi image dikhti hai; note mein likha hai real, inverted, badi. Step 5: ab concave lens; ek parallel ray aise diverge hoti hai jaise object side ke F se aa rahi ho (F tak dashed peeche), aur centre se guzarti ray seedhi aage jaati hai; peeche trace karne par woh F aur lens ke beech ek chhoti, seedhi, dashed image par milti hain.

Lens ka magnification bilkul mirror ki tarah define hota hai, aur O se guzarte similar triangles dete hain

lens ka magnification

m = \frac{h^{'}}{h} = \frac{v}{u}

(9.24)

Real image ke liye $u < 0$ aur $v > 0$ , isliye $m < 0$ : image inverted hai. Lens se bani virtual image object side par hoti hai, $v < 0$ , isliye $m > 0$ aur image erect hai. Neeche lens bench par dono lenses try karo aur signed readouts se $1 / v - 1 / u = 1 / f$ check karo.

InteractiveLens bench

Ek interactive ray diagram. Do buttons se convex ya concave lens chuna jaata hai. Sliders object distance ko 4 se 60 centimetre aur focal length ko 5 se 25 centimetre tak set karte hain. Drawing mein lens, uske do foci aur focal length ke double wale points, 2 centimetre lamba object, parallel ray, centre se guzarti ray aur focus se guzarti ray dikhti hai, virtual image hone par dashed backward extensions ke saath, aur khud image bhi. Readouts signed object distance, image distance, focal length, magnification aur image height dete hain; ek status line image ko describe karti hai.

Fig. 9.19Convex ya concave lens chuno, phir object ko move karo aur focal length badlo. u, v, f aur m ki signed values padho.

Example 9.6NCERT Example 9.6 · p. 236

Ek lens, teen liquids

Ek thin double convex lens $n = 1.52$ wale glass se bana hai. Use baari-baari se teen liquids mein duboya jaata hai, jinka index (a) $1.33$ , (b) $1.52$ aur (c) $1.63$ hai. Air ke muqable har liquid mein yeh kaisa behave karta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.5.3 Lens ki power

Lens light ko kitna zor se modta hai, yeh uski power se naapa jaata hai. Height $h$ par axis ke parallel ray angle $δ$ se mudti hai taaki woh focus par axis ko kaate: $\tan δ = h / f$ . Unit height par ray aur chhote angles ke liye $δ \approx 1 / f$ : focal length jitni chhoti, lens beam ko utna tezi se converge (ya negative $f$ ke liye diverge) karta hai. Isliye hum lens ki power (लेंस की क्षमता) ko aise define karte hain

lens ki power

P = \frac{1}{f}

(9.25)

Definition

Lens ki power

Metres mein focal length ka reciprocal, $P = 1 / f$ . Iski SI unit dioptre (D) hai: $1 D = 1 m^{- 1}$ . Converging lenses ki power positive aur diverging lenses ki power negative hoti hai.

Example $+ 0.25 m$ focal length wale lens ki power $+ 4 D$ hai.

Opticians prescriptions dioptres mein likhte hain. $+ 2.0 D$ ke reading glasses $f = + 50 c m$ wale converging lenses hain; short-sighted insaan ke $- 2.5 D$ ke glasses $f = - 40 c m$ wale diverging lenses hain.

Try karke dekho

Reading glasses ki power nikaalo

Tumhe chahiye

reading glasses (ya koi bhi magnifying glass)
white paper ki ek sheet
ek metre scale
ek window jiske bahar bright view ho

Pehle guess karo

Reading glasses par

+ 2.5

likha hai. Door ki window ki sharp image banane ke liye lens ko paper se lagbhag kitni door pakadna hoga?

lagbhag 2.5 cmlagbhag 25 cmlagbhag 40 cmlagbhag 2.5 m

Window se kamre ki doosri taraf khade ho jao aur paper ko window ki taraf karke pakdo.
Glasses ka ek lens window aur paper ke beech pakdo aur use tab tak khiskao jab tak window frame ki image sharp na ho jaaye.
Lens se paper tak ki distance naapo: yeh (lagbhag) focal length $f$ hai, kyunki window door hai.
$f$ ko metres mein lekar $P = 1 / f$ nikaalo, aur frame par chhape number ya prescription se compare karo.

Table 9.4Tumhare measurements

Lens	f (m)	P = 1/f (D)
left lens
right lens

Tumhari entries apne aap save hoti hain.

Example 9.7NCERT Example 9.7 · p. 236

Power, refractive index aur paani ke neeche lens

(a) $f = - 25 c m$ wale diverging lens ki power kya hai? (b) Ek double convex lens ke faces ke radii $12 c m$ aur $18 c m$ hain aur air mein focal length $12 c m$ hai. Uske glass ka refractive index nikaalo. (c) Ek glass lens ( $n = 1.60$ ) ki air mein focal length $15 c m$ hai. Water ( $n = 1.33$ ) mein uski focal length kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.5.4 Contact mein rakhe thin lenses ka combination

Camera aur microscope ke objectives kai lenses ko aapas mein chipka kar banaye jaate hain. Contact mein rakhe $f_{1}$ aur $f_{2}$ focal length wale do thin lenses ka optical centre lagbhag ek hi hota hai. Kaunsa single lens yahi kaam karega?

DerivationContact mein lenses ki focal length

Object $u$ par ho. Pehla lens akele $v_{1}$ par image banata:

\frac{1}{v_{1}} - \frac{1}{u} = \frac{1}{f_{1}}

(9.27)

Doosra lens us point ki taraf converge hoti (ya usse diverge hoti) light leta hai, isliye $v_{1}$ uski object distance hai. Yeh final image $v$ par banata hai:

\frac{1}{v} - \frac{1}{v_{1}} = \frac{1}{f_{2}}

(9.28)

Pehli image calculation ka bas ek step hai; light kabhi asal mein wahan pahunchti nahi, kyunki doosra lens beech mein hai. Dono equations jodne par,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{1}} + \frac{1}{f_{2}}

(9.29)

Yeh $f$ focal length wale single lens ka lens formula hai, jahan

contact mein lenses

\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}

(9.30)

Zyada lenses ke liye yahi argument dohraane se aur terms judte hain:

\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} + \dots

(9.31)

powers jud jaati hain

P = P_{1} + P_{2} + P_{3} + \dots

(9.32)

Yeh sum algebraic hai: diverging lenses negative powers jodte hain. Aur kyunki har image agle lens ka object hai, overall magnification alag-alag magnifications ka product hai:

m = m_{1} m_{2} m_{3} \dots

(9.33)

Lenses ko jodne se designer aisi power tak pahunch sakta hai jo kisi single lens mein nahi, aur colours ko alag tarah phailane wale glasses ka pair banakar coloured fringes jaise defects cancel kar sakta hai. Is tarah ke lens systems har camera, microscope aur telescope ke andar hote hain.

Example

Contact mein do lenses

$+ 4.0 D$ power ka ek converging lens $- 1.5 D$ power ke diverging lens ke contact mein rakha hai. Combination ki power aur focal length nikaalo, aur batao ki yeh light ko converge karta hai ya diverge.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9.8NCERT Example 9.8 · p. 238

Alag-alag rakhe do lenses

Ek object $20 c m$ focal length wale converging lens ke $30 c m$ saamne khada hai. $40 c m$ focal length ka ek diverging lens pehle lens ke $30 c m$ peeche, same axis par rakha hai. Final image ki position, nature aur object ke muqable size nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek double convex lens ke radii

20 c m

aur

30 c m

hain aur woh

n = 1.50

wale glass se bana hai. Air mein uski focal length kitni hai, cm mein?

Answer dekho

Answer 24 cm

1 / f = 0.50 (1 / 20 + 1 / 30) = 0.50 \times 5 / 60 = 1 / 24

, isliye

f = 24 c m

. Signs ke saath

R_{1} = + 20

aur

R_{2} = - 30

, aur

- 1 / R_{2}

judta hai.

9.6 Prism se refraction

Triangular glass prism ke do refracting faces hote hain jo angle of the prism $A$ par milte hain. Ek ray pehle face mein angle of incidence $i$ par ghusti hai aur $r_{1}$ par refract hoti hai, glass paar karti hai, doosre face se $r_{2}$ par milti hai aur angle of emergence $e$ par nikalti hai. Slab ke ulat, yahan faces parallel nahi hain, isliye dono baar ka mudna cancel hone ki jagah jud jaata hai: ray base ki taraf angle of deviation $δ$ se mud kar nikalti hai (Fig. 9.20).

3/4Incident direction aur emergent ray ke beech ka angle deviation $δ$ hai.

Fig. 9.20Dono refractions ray ko base ki taraf modte hain; kul mod deviation

δ

hai.NCERT Fig. 9.21 · p. 239

Ek equilateral glass prism jiska apex angle A top par hai. Step 1: left se aati ek ray left face se dashed normal ke saath angle i par milti hai aur glass mein chhote angle r1 par refract hoti hai. Step 2: ray prism paar karti hai, right face se uske normal ke saath angle r2 par milti hai aur bade angle e par nikalti hai, normal se door aur base ki taraf mud kar. Step 3: incident ray ko dashed line ke roop mein seedha aage badhaya gaya hai, aur emergent ray ko peeche trace kiya gaya hai; unke beech ka angle, delta, deviation hai. Step 4: orange rang ki doosri ray bade angle par ghusti hai taaki woh prism ko base ke parallel paar kare aur usi angle par nikle jis par ghusi thi; yeh minimum deviation ka raasta hai.

DerivationPrism mein angles

Jin points par ray glass mein ghusti aur nikalti hai unhe E aur X kaho, aur wahan normals banao; woh prism ke andar ek point O par milte hain. Har normal apne face ke perpendicular hai, isliye dono normals usi angle par kaatte hain jis par dono faces: O par triangle EOX ka exterior angle $A$ ke barabar hai. Us triangle ke E aur X par interior angles $r_{1}$ aur $r_{2}$ hain, aur triangle ka exterior angle saamne ke do interior angles ke sum ke barabar hota hai. Isliye

r_{1} + r_{2} = A

(9.34)

Pehle face par ray $i - r_{1}$ mudti hai; doosre par, usi taraf, $e - r_{2}$ . Kul deviation inka sum hai:

prism se deviation

δ = (i - r_{1}) + (e - r_{2}) = i + e - A

(9.35)

Isliye deviation is par depend karta hai ki ray kaise ghusti hai. Jaise-jaise $i$ chhoti values se badhta hai, $δ$ pehle ghatta hai, ek minimum tak pahunchta hai, aur phir badhta hai. Neeche prism lab se curve explore karo. Curve ke bilkul neeche ko chhod kar, har deviation do angles of incidence par hota hai. Yeh sanyog nahi hai: $i$ aur $e$ badalne par (9.35) nahi badalta, aur unhe badalna bas light ko usi raaste par ulta chalana hai, jo same deviation deta hai.

InteractivePrism lab

Ek interactive prism. Upar ki drawing mein ek ray prism ke left face mein ghusti aur right face se nikalti dikhti hai, dashed normals aur incident ray ke dashed continuation ke saath; agar ray doosre face par totally reflect ho, ya base tak pahunche, to drawing aur ek message yeh batate hain. Uske neeche ek graph abhi ke prism ke liye 0 se 90 degrees tak angle of incidence ke against deviation plot karta hai, abhi ki ray ke liye ek dot aur minimum deviation par ek dashed line ke saath. Sliders angle of incidence, 30 se 75 degrees tak prism angle aur 1.30 se 1.80 tak refractive index set karte hain, aur ek button angle of minimum deviation par le jaata hai. Readouts r1, r2, e, deviation, minimum deviation aur woh incidence dete hain jis par yeh hota hai.

Fig. 9.21Angle of incidence, prism angle aur refractive index badlo. Graph par dot abhi ka deviation dikhata hai; dashed line minimum mark karti hai.

Curve ke bilkul neeche sirf ek $i$ yeh deviation deta hai, isliye raasta khud apna ulta hona chahiye: $e = i$ aur $r_{1} = r_{2}$ . Wahan ka deviation angle of minimum deviation $D_{m}$ hai, aur barabar sides wale prism ke liye andar ki ray base ke parallel chalti hai. (9.34) aur (9.35) mein $r_{1} = r_{2} = r$ aur $δ = D_{m}$ , $i = e$ rakhne par:

r = \frac{A}{2}, i = \frac{A + D_{m}}{2}

(9.37)

minimum deviation se refractive index

n_{21} = \frac{n_{2}}{n_{1}} = \frac{\sin [(A + D_{m}) / 2]}{\sin (A / 2)}

(9.38)

$A$ aur $D_{m}$ dono precisely naape ja sakte hain, jaise spectrometer se, isliye (9.38) kisi glass ka, aur har colour ka alag se, refractive index nikaalne ka standard laboratory tareeka hai.

Thin prism ke liye, jiska $A$ bas kuch degrees hai, $D_{m}$ bhi chhota hota hai, aur sines ki jagah unke angles rakhe ja sakte hain: $n_{21} \approx \frac{(A + D_{m}) / 2}{A / 2}$ , jo deta hai

D_{m} \approx (n_{21} - 1) A

$n = 1.5$ wale glass ka $4^{\circ}$ prism light ko sirf lagbhag $2^{\circ}$ deviate karta hai. Jin logon ki aankhein ek saath ek taraf point nahi karti, unke spectacles mein kabhi-kabhi aise kamzor prisms hote hain.

Example

Minimum deviation se refractive index

Light $60^{\circ}$ angle wale glass prism se guzarti hai. Angle of minimum deviation $46^{\circ}$ naapa gaya. Glass ka refractive index, minimum deviation par angle of incidence, aur usi glass ke $5^{\circ}$ prism se hone wala deviation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek ray

60^{\circ}

prism se

i = 50^{\circ}

aur

e = 40^{\circ}

ke saath guzarti hai. Deviation kitna hai, degrees mein?

Answer dekho

Answer 30 °

δ = i + e - A = 50^{\circ} + 40^{\circ} - 60^{\circ} = 30^{\circ}

. Yahi deviation

i = 40^{\circ}

(ulta raasta) par bhi hoga.

9.7 Optical instruments

Mirrors, lenses aur prisms parts hain; optical instruments unhe jodte hain. Periscope deewar ke upar dekhne ke liye do mirrors ya totally reflecting prisms use karta hai; kaleidoscope mirrors ke beech baar-baar reflections se patterns banata hai; binoculars lenses aur prisms ko jodte hain. Sabse important optical instrument insaan ki aankh hai, jo tumne Class 10 mein padhi thi. Yahan hum un do instruments ko dekhte hain jinhone bahut chhoti aur bahut door ki duniya khol di: microscope aur telescope.

9.7.1 Microscope

Koi cheez kitni badi dikhti hai, yeh uske asli size par nahi, balki is par depend karta hai ki woh tumhari aankh par kitna angle banati hai: haath bhar door rakha coin Moon ko chhupa sakta hai. Chhoti cheez ko behtar dekhne ke liye tum use paas laate ho, lekin aankh near point se paas ki kisi cheez par focus nahi kar sakti, jo normal adult ke liye lagbhag $D = 25 c m$ hai. Isliye height $h$ wali chhoti cheez nangi aankh se aaraam se dekhne par sabse bada angle lagbhag $h / D$ bana sakti hai.

Simple microscope, yaani magnifying glass, chhoti focal length wala converging lens hai jo aankh ke paas pakda jaata hai. Object ko focal point par ya usse thoda andar rakha jaata hai. Tab lens ek erect, enlarged, virtual image itni door banata hai ki aankh us par focus kar sake (Fig. 9.22).

2/3F par object: rays $θ_{i} \approx h / f$ par parallel nikalti hain, aur relaxed aankh image ko infinity par dekhti hai.

Fig. 9.22Magnifying glass object ko 25 cm se paas aane deta hai jabki uski image aaraam wali distance par rehti hai.NCERT Fig. 9.23 · p. 241

Right mein ek aankh axis ke along dekhti hai. Step 1: bina lens ke, ek chhoti cheez aankh se 25 centimetre door, near point par khadi hai; uske top se aankh tak ki line axis ke saath chhota angle theta zero banati hai. Step 2: aankh ke theek saamne ek convex lens rakha jaata hai aur object ko uske focus F par le jaaya jaata hai; object ke top se do rays, ek axis ke parallel aur ek lens ke centre se, lens se ek doosre ke parallel bade angle theta i par nikalti hain, isliye image infinity par hai aur relaxed aankh use dekh sakti hai. Step 3: object ko F se thoda andar khiskaya jaata hai; lens se nikalti rays ab diverge hoti hain, aur unke dashed backward extensions aankh se 25 centimetre door ek badi, seedhi, virtual image par milte hain, jo object se chhe guna lambi hai.

DerivationSimple microscope ki magnifying power

Near point par image. Lens formula $1 / v - 1 / u = 1 / f$ ko $v$ se multiply karo: $1 - v / u = v / f$ , isliye $m = v / u = 1 - v / f$ . Virtual image object side par near point par hai, $v = - D$ :

near point par image

m = 1 + \frac{D}{f}

(9.39)

$D = 25 c m$ ke saath, $5 c m$ focal length wala lens $m = 6$ deta hai. Kyunki image theek $D$ par hai, yahi number uske angular size ko near point par dekhe gaye object ke angular size se compare karta hai.

Infinity par image. Near point par image dekhne se aankh par zor padta hai, isliye zyada aaraam wale arrangement mein object focal point par rakha jaata hai. Ab useful measure magnifying power, yaani angular magnification, hai. Lens ke bina sabse achha angle hai

θ_{0} \approx \tan θ_{0} = \frac{h}{D}

(9.40)

Aankh lens ke paas aur object focus par ho, to object ke top se rays parallel nikalti hain, axis se itne jhuki hui:

θ_{i} \approx \frac{h}{f}

(9.41)

Isliye angular magnification hai

infinity par image

m = \frac{θ_{i}}{θ_{0}} = \frac{D}{f}

(9.42)

Yeh (9.39) se ek kam hai, relaxed viewing ke liye chhoti si keemat; ab se hum final image infinity par lenge. Dekho ki lens asal mein kya karta hai. Aankh lens par ho, to virtual image utna hi angle banati hai jitna object khud. Fayda yeh hai ki object aankh se kuch centimetre tak laaya ja sakta hai, jahan woh bada angle banata hai, jabki uski image aisi distance par rehti hai jahan aankh focus kar sake.

Simple microscope lagbhag $9$ ya $10$ guna tak hi limited hai. Chhoti focal lengths ke liye chhote, bahut curved lenses chahiye jinki images edges par dhundhli aur rangeen hoti hain. Isse zyada ke liye hum do lenses ka magnification compound karte hain.

Compound microscope kaam ko do lenses mein baant deta hai (Fig. 9.23). Objective, jo specimen ke bahut paas hota hai, ki focal length bahut chhoti hoti hai; specimen uske focus se thoda bahar ho, to woh tube mein beech tak ek real, ulti aur pehle se badi picture banata hai. Phir eyepiece us picture ke saath waisa hi karta hai jaisa magnifying glass postage stamp ke saath. Picture ko eyepiece ke focus par (ya usse thoda andar) giraaya jaata hai, aur aankh use eyepiece se dekhti hai, aur final image infinity (ya near point) par dikhti hai. Kyunki koi use wapas seedha nahi karta, specimen ulta dikhta hai, jo cells ki slide ke liye shayad hi maayne rakhta hai.

2/3First image eyepiece ke focus par hai; eyepiece rays ko parallel bahar bhejta hai, isliye final image infinity par hai.

Fig. 9.23Objective linearly magnify karta hai; phir eyepiece angularly magnify karta hai.NCERT Fig. 9.24 · p. 243

Left mein ek chhota objective lens hai, jiske focus se theek aage ek bahut chhota object hai, aur right mein ek bada eyepiece lens. Step 1: object ke top se do rays, ek axis ke parallel aur ek objective ke centre se, lenses ke beech milkar ek badi, ulti, real first image banati hain. Step 2: is first image ke top se, jo eyepiece ke focus par hai, axis ke parallel ek ray aur eyepiece ke centre se ek ray eyepiece se ek doosre ke parallel nikalti hain, isliye final image infinity par hai. Step 3: lengths mark ki gayi hain: f o, objective ki focal length; L, objective ke second focus se eyepiece ke first focus tak tube length; aur f e, eyepiece ki focal length.

Objective ka magnification. Us ray ko follow karo jo object ke tip se axis ke parallel nikalti hai. Objective ke baad woh second focal point se guzarti hai aur first image ke tip tak jaati hai. Yeh poore raaste axis ke saath ek hi angle $ϕ$ banati hai, isliye focal point se pehle bana right triangle (height $h$ , base $f_{o}$ ) aur uske baad wala (height $h^{'}$ , base $L$ ) similar hain: $\tan ϕ = h / f_{o} = h^{'} / L$ . Isliye

objective

m_{o} = \frac{h^{'}}{h} = \frac{L}{f_{o}}

(9.43)

Yahan $L$ , yaani tube length, objective ke second focal point se eyepiece ke first focal point tak ki distance hai. Eyepiece ek simple microscope hai, isliye uska angular magnification hai

final image near point par

m_{e} = 1 + \frac{D}{f_{e}}

(9.44a)

final image infinity par

m_{e} = \frac{D}{f_{e}}

(9.44b)

(9.33) se total magnification product hai. Final image infinity par ho to:

compound microscope

m = m_{o} m_{e} = \frac{L}{f_{o}} \cdot \frac{D}{f_{e}}

(9.45)

Bade magnification ke liye dono focal lengths chhoti honi chahiye. $1 c m$ se kaafi kam focal length wale lenses achhe se banana mushkil hai, aur lambi tube instrument ko bhaari-bharkam bana deti hai. Modern objectives aur eyepieces mein kai lenses hote hain jo ek doosre ke defects cancel karte hain, aur specimen par achhi roshni magnification jitni hi important hai.

ExampleNCERT p. 244

School ka microscope

Ek compound microscope ke objective ki focal length $0.80 c m$ , eyepiece ki focal length $2.5 c m$ aur tube length $16 c m$ hai. Final image infinity par ho tab, aur final image near point par ho tab, iski magnifying power nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.7.2 Telescope

Telescope door ki cheezon ki directions ke beech ke angle ko magnify karta hai. Refracting telescope mein objective lambi focal length aur chaude aperture wala lens hota hai. Door ki cheez se light axis se chhote angle $α$ par jhuki parallel rays ke roop mein aati hai, aur objective unhe apne focal plane mein ek real, inverted image par laata hai. Us image se ek focal length aage rakha eyepiece rays ko phir parallel bahar bhejta hai, ab bade angle $β$ par jhuki hui (Fig. 9.24).

3/3Eyepiece light ko bade angle $β$ par parallel bahar bhejta hai; tube $f_{o} + f_{e}$ lambi hai.

Fig. 9.24Telescope chhote angle α ko bade angle β mein badalta hai, jahan β/α = fo/fe.NCERT Fig. 9.25 · p. 245

Left mein ek bada objective lens aur right mein ek chhota eyepiece lens, same axis par. Step 1: door ki cheez se teen parallel rays axis se chhote angle alpha par aati hain. Step 2: objective unhe apne focal plane mein axis ke neeche ek point par milata hai, jahan ek chhoti ulti real image banti hai. Step 3: rays eyepiece se hokar aage jaati hain, jo image se ek eyepiece focal length aage hai, aur ek doosre ke parallel bade angle beta par nikalti hain; objective se image tak distance f o aur image se eyepiece tak f e mark ki gayi hai.

Agar first image ki height $h$ hai, to objective use angle $α \approx h / f_{o}$ par dekhta hai, aur eyepiece uske tip se rays ko $β \approx h / f_{e}$ par bahar bhejta hai. Magnifying power hai

telescope

m = \frac{β}{α} \approx \frac{h / f_{e}}{h / f_{o}} = \frac{f_{o}}{f_{e}}

(9.46)

Is normal adjustment mein tube length $f_{o} + f_{e}$ hai. $f_{o} = 90 c m$ aur $f_{e} = 3.0 c m$ wala telescope $30$ guna magnify karta hai: aasmaan mein ${0.2}^{\circ}$ door do taare $6^{\circ}$ door dikhte hain. Astronomical telescope cheezein ulti dikhata hai, jo taaron ke liye maayne nahi rakhta; terrestrial telescope erecting lenses ka ek pair jodta hai, aur binoculars totally reflecting prisms se image ko seedha karte hain aur lambe light path ko chhote case mein mod dete hain.

Example

Ek chhota refracting telescope

Ek refracting telescope ke objective ki focal length $90 c m$ aur eyepiece ki focal length $3.0 c m$ hai. Normal adjustment mein iski magnifying power aur tube length nikaalo. Earth se dekhne par Jupiter ke do moons ${0.10}^{\circ}$ door hain. Telescope se woh kitne door dikhte hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Astronomers ke liye magnification sasta hissa hai: alag eyepiece lagao aur woh badal jaata hai. Do cheezein is tarah nahi khareedi ja sakti. Telescope apne poore objective par starlight jama karta hai, isliye diameter double karne se chaar guna light jama hoti hai aur kaafi dhundhli cheezein dikhne lagti hain: uski light-gathering power badhti hai. Chauda objective lagbhag ek hi direction mein pade do taaron ko ek blur ki jagah do points ki tarah dikhata hai: uski resolving power behtar hoti hai. Isliye hamesha bade objectives ki race rahi hai. Lenses yeh race haar jaate hain. Bada glass disc bhaari hota hai, sirf apne kinare se pakda ja sakta hai (jahan woh latak jaata hai), aur apni poori motaai mein bina kisi flaw aur colour defect ke hona chahiye. Ab tak bana sabse bada lens telescope, jo 1897 mein USA ki Yerkes Observatory mein poora hua, uske objective ka diameter lagbhag $1 m$ (40 inches) hai.

Isliye lagbhag har bada research telescope ek reflecting telescope hai, jiska objective ek concave mirror hai; neeche ka comparison iski wajah batata hai. Mirror mein ek mushkil hai: woh light ko usi raaste wapas bhejta hai jis se woh aayi, isliye uska focus uske saamne, aati starlight ke raaste mein hota hai. Shuru ke bade telescopes ne bas astronomer ko wahin bitha diya. USA mein Mount Palomar ke $5.08 m$ (200-inch) telescope par observer prime focus par ek chhote cage mein baithta tha, jo mirror ka thoda sa hissa dhak deta tha. Cassegrain design zyada saaf hai. Ek chhota convex secondary mirror converging light ko focus tak pahunchne se pehle pakad leta hai aur use tube mein wapas neeche, primary ke beech ke ched se, uske peeche rakhe eyepiece ya camera tak bhejta hai (Fig. 9.25). Kyunki secondary effective focal length ko bhi badha deta hai, lambi focal length chhoti tube mein fit ho jaati hai.

Fig. 9.25Convex secondary light ko primary ke ched se wapas mod deta hai: chhoti tube mein lambi focal length.NCERT Fig. 9.26 · p. 246

Left mein khuli ek chhoti tube. Parallel rays andar aati hain aur right mein ek bade concave primary mirror tak jaati hain, jiske centre mein ched hai. Mirror unhe ek focus ki taraf reflect karta hai, lekin wahan pahunchne se pehle khule end ke paas ek chhota convex secondary mirror unhe wapas primary ki taraf reflect karta hai. Rays beech ke ched se guzarti hain aur primary ke peeche image par milti hain, jahan eyepiece rakha hai.

Compare

Refracting aur reflecting telescopes

Refracting (objective ek lens)

light glass ke through jaati hai, jo poora flawless hona chahiye
rangeen fringes (chromatic aberration)
bhaari lens sirf kinare se pakda jaata hai
practical limit lagbhag 1 m

Reflecting (objective ek mirror)

sirf saamne ki surface perfect honi chahiye
koi chromatic aberration nahi
mirror ko poori peeth par support milta hai
8–10 m ke mirrors use mein hain

Chalo explore karein

Khilona telescope banao

Tumhe chahiye

+2.0 D ka spectacle lens ya lagbhag 50 cm focal length ka magnifier
lagbhag 5 cm focal length ka magnifying glass
cardboard ki do tubes jo ek doosre ke andar khisakti hon
tape

Pehle guess karo

In do lenses ke saath tum lagbhag kitni magnifying power expect karte ho?

lagbhag 2.5×lagbhag 10×lagbhag 55×

Kamzor lens (lambi focal length) ko chaudi tube ke ek end par tape karo: yeh objective hai.
Strong lens ko patli tube ke ek end par tape karo: yeh eyepiece hai.
Tubes ko tab tak khiskao jab tak lenses ke beech ki distance lagbhag $55 c m$ na ho, door ke ped ya building ko dekho, aur view sharp hone tak adjust karo.
Dhyaan do ki view ulta hai, aur lenses ke beech ki final distance naapo.

Quick check

4 mein se 1

Ek telescope ke objective ki focal length

1.2 m

aur eyepiece ki focal length

4.0 c m

hai. Normal adjustment mein iski magnifying power kya hai?

guna

Answer dekho

Answer 30 guna

m = f_{o} / f_{e} = 120 c m / 4.0 c m = 30

. Pehle dono focal lengths ko same unit mein lao.

Recap

Summary aur key points

Light electromagnetic waves ka visible band hai, $λ \approx 400$ – $750 n m$ ; kyunki $λ$ roz ki cheezon ke muqable bahut chhota hai, ise seedhi rays ke along chalta hua maana ja sakta hai (Section 9.1).
Reflection ke laws curved mirror ke har point par lagte hain, normal radius ke along hota hai. Cartesian convention: distances pole ya optical centre se, incident light ke along positive; upar ki heights positive (Section 9.2.1).
Paraxial rays ke liye spherical mirror ka $f = R / 2$ ; concave mirrors ka $f < 0$ , convex mirrors ka $f > 0$ (Section 9.2.2).
Signs use karne par mirror equation $1 / v + 1 / u = 1 / f$ aur magnification $m = h^{'} / h = - v / u$ har case mein sahi hain (Section 9.2.3).
Snell's law: $\sin i / \sin r = n_{21} = n_{2} / n_{1}$ , yaani $n_{1} \sin i = n_{2} \sin r$ . Slab ray ko deviate kiye bina side mein khiskata hai; lagbhag normal se dekhne par apparent depth = real depth / $n$ (Section 9.3).
Denser se rarer medium mein jaate hue, jab $i$ critical angle se zyada ho, to light totally reflect hoti hai, $\sin i_{c} = n_{2} / n_{1}$ (Section 9.4).
Totally reflecting $45^{\circ}$ prisms beams ko bina loss ke modte ya ulta karte hain; optical fibres light ko cladding se zyada index wale core ke along guide karte hain (Section 9.4.1).
Spherical refracting surface $n_{2} / v - n_{1} / u = (n_{2} - n_{1}) / R$ follow karti hai (Section 9.5.1).
Thin lens ke liye $1 / v - 1 / u = 1 / f$ ; lens maker's formula $1 / f = (n_{21} - 1) (1 / R_{1} - 1 / R_{2})$ hai, aur magnification $m = v / u$ (Section 9.5.2).
Power $P = 1 / f$ dioptres mein; contact mein rakhe thin lenses ki powers judti hain aur magnifications multiply hote hain (Section 9.5.3, Section 9.5.4).
Prism mein $r_{1} + r_{2} = A$ aur $δ = i + e - A$ ; minimum deviation par $i = e$ aur $n_{21} = \sin [(A + D_{m}) / 2] / \sin (A / 2)$ ; thin prism $(n - 1) A$ deviate karta hai (Section 9.6).
Simple microscope $1 + D / f$ (image $D = 25 c m$ par) ya $D / f$ (image infinity par) magnify karta hai; compound microscope lagbhag $(L / f_{o}) (D / f_{e})$ magnify karta hai (Section 9.7.1).
Normal adjustment mein telescope ka angular magnification $f_{o} / f_{e}$ aur length $f_{o} + f_{e}$ hoti hai; bade telescopes chromatic aberration se bachne aur bade objectives ko support karne ke liye mirrors use karte hain (Section 9.7.2).

Flash cards · 12

Formula revision

1 / 12

Practice · 36 questions

Practice questions answers ke saath (NCERT “Exercises”)

1
NumericalEasyNCERT Exercises 9.1
Physics lab mein $40 c m$ radius of curvature wale concave mirror se ek jalti LED, jo $3.0 c m$ lambi hai, ki image white card par daali jaati hai. LED mirror se $30 c m$ door hai.
1. Sharp picture ke liye card kahan pakadna hoga?
  cm
  Solution dikhao
  $f = - 20 c m$ , $u = - 30 c m$ : $1 / v = 1 / f - 1 / u = - 1 / 20 + 1 / 30 = - 1 / 60$ , isliye $v = - 60 c m$ : mirror ke $60 c m$ saamne.
2. Picture kitni lambi hai, aur kis taraf se seedhi hai?
  Solution dikhao
  $m = - v / u = - (- 60) / (- 30) = - 2$ : real, inverted aur $6.0 c m$ lambi.
3. Ab LED ko thoda-thoda karke mirror ki taraf khiskaya jaata hai. Picture sharp rakhne ke liye card ko kaise khiskana hoga, aur jab LED $20 c m$ se paas aa jaaye to kya hota hai?
  Solution dikhao
  Jaise-jaise LED F ( $20 c m$ ) ke paas aati hai, image distance bina kisi limit ke badhti hai, isliye card ko aur aur peeche le jaana padta hai. F ke andar image virtual ho jaati hai, mirror ke peeche, aur card ki koi position use pakad nahi sakti.
2
NumericalEasyNCERT Exercises 9.2
Ek bachcha $2.0 c m$ lambi toy car ko polished steel ki kadchhi ki peeth ke $30 c m$ saamne pakadta hai, jo $20 c m$ focal length wale convex mirror ki tarah kaam karti hai.
1. Toy ki image kahan hai, aur woh kitni lambi hai?
  cm
  Solution dikhao
  $f = + 20 c m$ , $u = - 30 c m$ : $1 / v = 1 / 20 + 1 / 30 = 1 / 12$ , isliye $v = + 12 c m$ , kadchhi ke peeche.
  $m = - v / u = - 12 / (- 30) = + 0.40$ : virtual, erect, $0.80 c m$ lambi.
2. Bachcha toy ko dheere-dheere aur door le jaata hai. Image ka kya hota hai?
  Woh F ki taraf jaati hai aur chhoti hoti hai, virtual hi rehti haiWoh F se aage jaati hai aur badi hoti haiToy kaafi door ho jaaye to woh real ho jaati haiWoh same size ki rehti hai
  Solution dikhao
  Jab $u \to - \infty$ , to $1 / v \to 1 / f$ : image mirror ke $20 c m$ peeche F ki taraf khisakti hai aur chhoti hoti jaati hai, lekin virtual aur erect rehti hai.
3
NumericalMediumNCERT Exercises 9.3
Ek beaker ke tal par $24.0 c m$ paani ( $n = 1.33$ ) ke neeche ek coin pada hai. Ek travelling microscope upar se use dekhta hai.
1. Microscope ko coin water surface se kitni gehrayi par dikhta hai?
  cm
  Solution dikhao
  Surface se $24.0 / 1.33 = 18.0 c m$ neeche.
2. Paani ki jagah utni hi gehrayi tak ek unknown liquid bhara jaata hai. Coin par dobara focus karne ke liye microscope ko $3.0 c m$ upar uthana padta hai. Liquid ka refractive index nikaalo.
  (koi unit nahi)
  Solution dikhao
  Coin ab $3.0 c m$ upar, $15.0 c m$ ki apparent depth par dikhta hai. Isliye $n = 24.0 / 15.0 = 1.60$ .
NumericalMediumNCERT Exercises 9.4
Air se glass block mein normal se $50^{\circ}$ par ghusti ray ${29.0}^{\circ}$ par refract hoti hai. Air se water mein $50^{\circ}$ par ghusti ray ${35.2}^{\circ}$ par refract hoti hai. Ek aquarium ka floor glass ka hai; uske water mein chalti ek ray glass se normal ke saath $40^{\circ}$ par milti hai. Glass mein woh kis angle par chalti hai?
°
Solution dikhao
$n_{g} = \sin 50^{\circ} / \sin {29.0}^{\circ} = 0.766 / 0.485 = 1.58$ ; $n_{w} = \sin 50^{\circ} / \sin {35.2}^{\circ} = 0.766 / 0.576 = 1.33$ .
Water–glass boundary par: $\sin r = (n_{w} / n_{g}) \sin 40^{\circ} = (1.33 / 1.58) (0.643) = 0.541$ , isliye $r = {32.7}^{\circ}$ , normal ki taraf thoda sa mudi hui.
5
NumericalMediumNCERT Exercises 9.5
Raat mein swimming pool ke upar mandraata ek drone pool ke floor par pade lamp ki photo leta hai. Water surface par $4.6 m$ chauda, sharp boundary wala bright circle dikhta hai, aur uske bahar andhera hai ( $n = 1.33$ ).
1. Bright patch ka edge sharp kyun hai?
  Solution dikhao
  Lamp ki jo light circle ke bahar surface tak pahunchti hai, woh critical angle ( ${48.8}^{\circ}$ ) se zyada angle par milti hai aur totally reflect hokar wapas pool mein chali jaati hai. Circle ke andar kuch light bahar refract hoti hai. 'Kuch nikalti hai' se 'kuch nahi nikalti' ka badlaav $i_{c}$ par achaanak hota hai.
2. Pool kitna gehra hai?
  m
  Solution dikhao
  $i_{c} = \sin^{- 1} (1 / 1.33) = {48.8}^{\circ}$ , $\tan i_{c} = 1.14$ . Radius $= 2.3 m$ , isliye depth $= 2.3 / 1.14 = 2.0 m$ .
6
NumericalMediumNCERT Exercises 9.6
Ek student air mein $60^{\circ}$ glass prism ka angle of minimum deviation $44^{\circ}$ naapti hai. Phir woh prism ko water ( $n = 1.33$ ) ke tank mein duba deti hai.
1. Uska measurement kaunsa refractive index deta hai?
  (koi unit nahi)
  Solution dikhao
  $n = \sin [(60^{\circ} + 44^{\circ}) / 2] / \sin 30^{\circ} = \sin 52^{\circ} / 0.500 = 1.58$ .
2. Water mein use kitne minimum deviation ki ummeed karni chahiye?
  °
  Solution dikhao
  Relative index $= 1.576 / 1.33 = 1.185$ . Phir $\sin [(60^{\circ} + D_{m}) / 2] = 1.185 \times 0.500 = 0.5925$ , isliye $(60^{\circ} + D_{m}) / 2 = {36.3}^{\circ}$ aur $D_{m} = {12.7}^{\circ}$ .
3. Index $1.58$ wale saaf oil mein prism light ke saath kya karega?
  Solution dikhao
  Relative index $1.576 / 1.58 \approx 1.00$ hai, isliye light mushkil se mudegi: prism lagbhag gayab ho jaayega.
NumericalEasyNCERT Exercises 9.7
Ek lens maker ko $15 c m$ focal length ke equiconvex lenses chahiye aur uske paas do glasses hain, refractive index $1.50$ aur $1.60$ wale. Har glass ke liye faces ka radius of curvature kitna hona chahiye, aur kaunsa glass zyada flat (patla) lens deta hai?
cm
Solution dikhao
$R = 2 (n - 1) f$ . $n = 1.50$ ke liye: $R = 2 \times 0.50 \times 15 = 15 c m$ . $n = 1.60$ ke liye: $R = 2 \times 0.60 \times 15 = 18 c m$ .
Zyada index wale glass ko kam curved faces ( $18 c m$ ) chahiye, isliye woh patla lens deta hai.
8
NumericalMediumNCERT Exercises 9.8
Ek projector lens beam ko screen ke ek point P ki taraf converge karta hai. P se $15 c m$ pehle beam mein ek doosra lens daala jaata hai.
1. Agar naya lens $f = 30 c m$ wala convex hai, to beam ab kahan converge hoti hai?
  cm
  Solution dikhao
  $1 / v = 1 / f + 1 / u = 1 / 30 + 1 / 15 = 1 / 10$ , isliye $v = + 10 c m$ : lens se $10 c m$ aage, P se paas.
2. Aur agar woh $f = 20 c m$ wala concave hai?
  cm
  Solution dikhao
  $1 / v = - 1 / 20 + 1 / 15 = 1 / 60$ , isliye $v = + 60 c m$ : ab bhi converging, lekin ab lens se $60 c m$ aage ke point par.
3. Usi jagah kaunsa lens rakhne se beam parallel nikalegi?
  Solution dikhao
  Parallel light ka matlab $v \to \infty$ : $1 / f = - 1 / u = - 1 / 15$ , isliye $15 c m$ focal length wala diverging lens ( $f = - 15 c m$ ).
NumericalEasyNCERT Exercises 9.9
$f = - 15 c m$ wale diverging lens se usse $30 c m$ door rakhe $2.0 c m$ oonche object ko dekha jaata hai. Image ko poori tarah describe karo. Kya object ko khiskane se is lens ki image kabhi real, ya object se badi, ho sakti hai?
cm
Solution dikhao
$1 / v = 1 / f + 1 / u = - 1 / 15 - 1 / 30 = - 1 / 10$ , isliye $v = - 10 c m$ : object wali side par hi.
$m = v / u = (- 10) / (- 30) = + 1 / 3$ : virtual, erect, $0.67 c m$ oonchi.
Kisi bhi real object ke liye $1 / f + 1 / u$ ke dono terms negative hain, isliye $v$ hamesha negative (virtual) hai aur $∣ v ∣ < ∣ u ∣$ , jisse $0 < m < 1$ milta hai. Object ko door le jaane se image bas F ki taraf jaati hai aur chhoti hoti hai.
NumericalEasyNCERT Exercises 9.10
Do thin lenses ek doosre ko chhoote hain: $f = + 25 c m$ wala converging aur $f = - 40 c m$ wala diverging. Pair ki focal length nikaalo. Kya yeh light ko converge karta hai ya diverge?
cm
Solution dikhao
$1 / f = 1 / 25 - 1 / 40 = (8 - 5) / 200 = 3 / 200$ , isliye $f = + 66.7 c m$ .
Positive: pair ek kamzor converging lens hai, kyunki dono mein converging lens zyada strong hai.
11
NumericalHardNCERT Exercises 9.11
Ek microscope ka objective ( $f = 1.5 c m$ ) aur eyepiece ( $f = 5.0 c m$ ) $16 c m$ door fixed hain, aur slide ko khiska kar focus kiya jaata hai. Ek student pehle relaxed viewing (final image infinity par) ke liye focus karti hai, phir dobara focus karti hai taaki final image uske near point, $25 c m$ , par ho.
1. Har setting mein slide objective se kitni door hai?
  cm
  Solution dikhao
  Infinity: first image eyepiece ke focus par honi chahiye, objective ke $16 - 5.0 = 11.0 c m$ peeche. $1 / u_{o} = 1 / 11.0 - 1 / 1.5$ se $u_{o} = - 1.737 c m$ milta hai.
  Near point: $1 / u_{e} = - 1 / 25 - 1 / 5.0$ se $u_{e} = - 4.17 c m$ milta hai, isliye first image $16 - 4.17 = 11.83 c m$ par hai. $1 / u_{o} = 1 / 11.83 - 1 / 1.5$ se $u_{o} = - 1.718 c m$ milta hai.
  Dono settings ke beech slide objective ki taraf sirf lagbhag $0.19 m m$ khisakti hai.
2. Har setting mein magnifying power nikaalo.
  guna
  Solution dikhao
  Infinity: $∣ m_{o} ∣ = 11.0 / 1.737 = 6.33$ , $m_{e} = 25 / 5.0 = 5.0$ , isliye $m \approx 32$ .
  Near point: $∣ m_{o} ∣ = 11.83 / 1.718 = 6.89$ , $m_{e} = 1 + 25 / 5.0 = 6.0$ , isliye $m \approx 41$ .
12
NumericalHardNCERT Exercises 9.12
Ek student ke microscope mein objective ka $f = 6.0 m m$ aur eyepiece ka $f = 3.0 c m$ hai. Objective se $7.0 m m$ neeche ek cell sample sharp dikhta hai, aur final image near point ( $25 c m$ ) par hai.
1. Lenses ek doosre se kitni door hain, aur magnifying power kya hai?
  guna
  Solution dikhao
  Objective: $1 / v_{o} = 1 / 0.60 - 1 / 0.70 = 0.238 c m^{- 1}$ , isliye $v_{o} = 4.2 c m$ aur $∣ m_{o} ∣ = 4.2 / 0.70 = 6.0$ .
  Eyepiece: $1 / u_{e} = - 1 / 25 - 1 / 3.0$ , isliye $u_{e} = - 2.68 c m$ . Separation $= 4.2 + 2.68 = 6.9 c m$ .
  Magnifying power $= 6.0 \times (1 + 25 / 3.0) = 6.0 \times 9.33 = 56$ hai.
2. Iski jagah relaxed aankh se dekhne ke liye eyepiece ko andar khiskana chahiye ya bahar, aur kitna? Tab magnifying power kya hogi?
  guna
  Solution dikhao
  Final image infinity par ho, iske liye first image eyepiece ke focus par honi chahiye, isliye separation $4.2 + 3.0 = 7.2 c m$ ho jaata hai: eyepiece ko lagbhag $0.32 c m$ bahar khiskao.
  Magnifying power $= 6.0 \times 25 / 3.0 = 50$ hai.
NumericalEasyNCERT Exercises 9.13
Ek birdwatcher ka refracting telescope normal adjustment mein $124 c m$ lamba hai aur $30$ guna magnify karta hai. Uske objective aur eyepiece ki focal lengths nikaalo.
cm
Solution dikhao
$f_{o} = 30 f_{e}$ , isliye $31 f_{e} = 124 c m$ : $f_{e} = 4.0 c m$ aur $f_{o} = 120 c m$ .
14
NumericalMediumNCERT Exercises 9.14
Ek solar observatory $12 m$ focal length wala lens objective use karti hai.
1. $2.0 c m$ eyepiece ke saath yeh kitna angular magnification dega?
  guna
  Solution dikhao
  $m = f_{o} / f_{e} = 1200 c m / 2.0 c m = 600$ .
2. Eyepiece ki jagah objective ke focus par ek card rakha jaata hai jo Sun ki image pakadta hai (koi usmein se dekhta nahi). Sun ka diameter $1.39 \times 10^{9} m$ aur distance $1.50 \times 10^{11} m$ hai. Sun ki image kitni chaudi hai?
  cm
  Solution dikhao
  Bana angle: $α = 1.39 \times 10^{9} / 1.50 \times 10^{11} = 9.27 \times 10^{- 3} r a d$ . Image width $= f_{o} α = 12 \times 9.27 \times 10^{- 3} = 0.111 m \approx 11 c m$ .
15
ProofMediumNCERT Exercises 9.15
Mirror equation $1 / v + 1 / u = 1 / f$ ko signs ke saath use karke, bina rays banaye, inmein se har ek dikhao.
1. C se aage rakhe object ki concave mirror F aur C ke beech real image banata hai.
  Solution dikhao
  Yahan $f < 0$ aur $u < 2 f$ (zyada negative). Tab $1 / u$ , $1 / (2 f)$ aur $0$ ke beech hai, isliye $1 / v = 1 / f - 1 / u$ , $1 / f$ aur $1 / (2 f)$ ke beech hai, dono negative. Isliye $v$ negative hai aur $f$ aur $2 f$ ke beech hai: F aur C ke beech ek real image.
2. Convex mirror real object ki hamesha virtual image banata hai.
  Solution dikhao
  $f > 0$ aur $u < 0$ , isliye $1 / v = 1 / f - 1 / u = 1 / f + 1 / ∣ u ∣ > 0$ : $v$ hamesha positive hai, mirror ke peeche. Image virtual hai.
3. Woh virtual image hamesha object se chhoti hoti hai aur pole aur focus ke beech hoti hai.
  Solution dikhao
  Kyunki $1 / v = 1 / f + 1 / ∣ u ∣$ , $1 / f$ aur $1 / ∣ u ∣$ dono se bada hai, image distance $v$ , $f$ aur $∣ u ∣$ dono se chhota hai. Isliye $0 < v < f$ : image P aur F ke beech hai. Aur $m = - v / u = v / ∣ u ∣$ , 0 aur 1 ke beech hai: erect aur diminished.
4. Pole aur focus ke beech rakhe object ki concave mirror virtual, enlarged image banata hai.
  Solution dikhao
  $f < 0$ aur $f < u < 0$ , isliye $∣ u ∣ < ∣ f ∣$ aur $- 1 / u = 1 / ∣ u ∣ > 1 / ∣ f ∣$ . Tab $1 / v = - 1 / ∣ f ∣ + 1 / ∣ u ∣ > 0$ , isliye $v > 0$ : virtual. Saath hi $1 / v < 1 / ∣ u ∣$ , isliye $v > ∣ u ∣$ aur $m = v / ∣ u ∣ > 1$ : enlarged aur erect.
16
NumericalMediumNCERT Exercises 9.16
Ek letter $12 c m$ mote flat slab jaise glass paperweight ( $n = 1.60$ ) ke neeche pada hai, aur tum use seedha neeche dekhte ho.
1. Letter kitna upar utha hua dikhta hai?
  cm
  Solution dikhao
  $12 c m$ glass sirf $12 / 1.60 = 7.5 c m$ mota dikhta hai, isliye letter $12 - 7.5 = 4.5 c m$ upar utha hua dikhta hai.
2. Ab wahi slab letter par rakhne ki jagah usse $10 c m$ upar pakda jaata hai. Ab letter kitna upar utha dikhta hai?
  cm
  Solution dikhao
  Ab bhi $4.5 c m$ . Sirf glass apparent distance ko $t (1 - 1 / n)$ se badalta hai; uske upar aur neeche ke air gaps jitne lambe hain utne hi dikhte hain. Isliye lagbhag normal se dekhne par shift is par depend nahi karta ki slab kahan hai.
17
NumericalHardNCERT Exercises 9.17
Medical endoscope ke ek fibre mein index $1.62$ ka glass core hai jis par index $1.50$ ke glass ki coating hai.
1. Air se flat end mein ghus kar core mein phansi rehne wali light fibre ki axis ke saath sabse chauda kitna angle bana sakti hai?
  °
  Solution dikhao
  $\sin i_{c} = 1.50 / 1.62 = 0.926$ , $i_{c} = {67.8}^{\circ}$ . Andar ray axis ke saath zyada se zyada $90^{\circ} - {67.8}^{\circ} = {22.2}^{\circ}$ bana sakti hai.
  End face par: $\sin θ = 1.62 \sin {22.2}^{\circ} = 0.612$ , isliye $θ = {37.7}^{\circ}$ . Axis se lagbhag $38^{\circ}$ ke andar ki light guide hoti hai.
2. Agar coating hata di jaaye aur core ke charon taraf air ho, to yeh kaise badlega?
  Solution dikhao
  Ab $\sin i_{c} = 1 / 1.62$ , $i_{c} = {38.1}^{\circ}$ , isliye andar axis ke saath angle ${51.9}^{\circ}$ tak ja sakta hai. Matching bahar ke angle ke liye $\sin θ = 1.62 \sin {51.9}^{\circ} = 1.27 > 1$ chahiye hoga, isliye end mein $90^{\circ}$ tak kisi bhi angle par ghusi light phans jaati hai. (Nanga fibre aasaani se kharonch jaata hai, aur use chhoone wali koi bhi cheez light ko bahar leak hone deti hai, isiliye fibres par coating hoti hai.)
NumericalMediumNCERT Exercises 9.18
Ek slide projector ko single convex lens se slide ki image slide se $2.4 m$ door screen par daalni hai. Lens ki focal length zyada se zyada kitni ho sakti hai?
m
Solution dikhao
Real image ke liye, jab object aur screen $D$ distance door hon, to $u$ aur $v$ ko $∣ u ∣ + v = D$ aur $1 / v + 1 / ∣ u ∣ = 1 / f$ satisfy karna hoga. $v$ hatane par $∣ u ∣$ mein ek quadratic milta hai jiske real solutions tabhi hain jab $D \geq 4 f$ .
Isliye $f_{\max} = D / 4 = 2.4 / 4 = 0.60 m$ ; tab slide aur screen dono lens se $1.2 m$ door hain ( $u = - 2 f$ , $v = 2 f$ ).
19
NumericalMediumNCERT Exercises 9.19
Ek lamp aur screen $80 c m$ door khade hain, aur $15 c m$ focal length wala convex lens unke beech bench par khiskaya jaata hai.
1. Lamp se kin distances par lens screen par sharp image deta hai?
  cm
  Solution dikhao
  $∣ u ∣ (80 - ∣ u ∣) = 15 \times 80 = 1200$ , isliye $∣ u ∣^{2} - 80 ∣ u ∣ + 1200 = 0$ , jisse $∣ u ∣ = 20 c m$ ya $60 c m$ milta hai. Dono positions $40 c m$ door hain.
2. Dono images ek doosre se kaise compare karti hain?
  Solution dikhao
  $∣ u ∣ = 20$ par $v = 60$ : $m = - 3$ . $∣ u ∣ = 60$ par $v = 20$ : $m = - 1 / 3$ . Dono real aur inverted hain; ek lamp se teen guna size ki hai, doosri ek-tihaai. Object aur image distances bas aapas mein badal jaati hain.
3. Dikhao ki general case mein, agar lamp–screen separation $D$ ke liye $d$ distance door do sharp positions hon, to $f = (D^{2} - d^{2}) / 4 D$ , aur yahan isse check karo.
  Solution dikhao
  Dono positions $∣ u ∣$ aur $v$ ko badal deti hain, isliye $∣ u ∣ + v = D$ aur $v - ∣ u ∣ = d$ , jisse $∣ u ∣ = (D - d) / 2$ , $v = (D + d) / 2$ milta hai. Tab $1 / f = 1 / v + 1 / ∣ u ∣ = 4 D / (D^{2} - d^{2})$ . Yahan $(6400 - 1600) / 320 = 15 c m$ .
20
NumericalHardNCERT Exercises 9.20
Question 10 ke dono lenses lo (converging, $f = 25 c m$ ; diverging, $f = 40 c m$ ) aur unhe common axis par $10 c m$ door lagao.
1. Pair se door ki ek street light dekhi jaati hai. Agar converging lens light ki taraf ho to uski image kahan hai? Aur agar diverging lens light ki taraf ho? Kya pair ko ek single focal length se describe kiya ja sakta hai?
  Solution dikhao
  Pehle converging lens: yeh apne $25 c m$ peeche focus karta, jo diverging lens se $15 c m$ aage hai, isliye $u_{2} = + 15 c m$ : $1 / v = - 1 / 40 + 1 / 15 = 1 / 24$ , diverging lens ke $v = 24 c m$ peeche.
  Pehle diverging lens: light uske $40 c m$ saamne, converging lens ke $50 c m$ saamne ke ek point se phailti hai: $1 / v = 1 / 25 - 1 / 50 = 1 / 50$ , isliye converging lens ke $v = 50 c m$ peeche.
  Light kis taraf jaati hai, us hisaab se focus aakhri lens se alag distance par hota hai, isliye fixed jagah par rakha koi single thin lens pair ki jagah nahi le sakta. (Combined focal length $1 / F = 1 / f_{1} + 1 / f_{2} - d / f_{1} f_{2}$ , jo $40 c m$ deti hai, tabhi hoti hai jab distances khaas reference planes se naapi jaayein.)
2. Ek $2.0 c m$ object converging lens ke $75 c m$ saamne khada hai. Final image ki position, nature aur size nikaalo.
  cm
  Solution dikhao
  Converging lens ke liye: $1 / v_{1} = 1 / 25 - 1 / 75 = 2 / 75$ , $v_{1} = 37.5 c m$ , $m_{1} = 37.5 / (- 75) = - 0.50$ .
  Diverging lens $10 c m$ aage hai, isliye $u_{2} = 37.5 - 10 = + 27.5 c m$ (virtual object): $1 / v_{2} = - 1 / 40 + 1 / 27.5$ , $v_{2} = 88 c m$ , $m_{2} = 88 / 27.5 = 3.2$ .
  $m = (- 0.50) (3.2) = - 1.6$ : real, inverted, $3.2 c m$ oonchi, diverging lens ke $88 c m$ peeche.
NumericalMediumNCERT Exercises 9.21
Index $1.50$ wale glass ke $60^{\circ}$ prism ke ek face mein light aise bhejni hai ki uska kuch hissa doosre face se nikal jaaye. Pehle face par sabse chhota kaunsa angle of incidence hai jis par koi bhi light bahar aati hai?
°
Solution dikhao
$\sin i_{c} = 1 / 1.50$ , $i_{c} = {41.8}^{\circ}$ . Limiting ray ke liye $r_{2} = {41.8}^{\circ}$ , isliye $r_{1} = 60^{\circ} - {41.8}^{\circ} = {18.2}^{\circ}$ .
$\sin i = 1.50 \sin {18.2}^{\circ} = 0.468$ , isliye $i = {27.9}^{\circ}$ . Isse chhote angles of incidence par light doosre face par andar hi totally reflect ho jaati hai; is angle par woh bas chhoo kar bahar nikalti hai.
22
NumericalMediumNCERT Exercises 9.22
Ek ghadi theek karne wala $2.0 m m$ door daanton wale gear ko $12 c m$ focal length wale loupe (magnifying lens) se dekhta hai, jo aankh par pakda hai, aur gear lens se $10 c m$ door hai.
1. Image kahan hai, aur usmein daante kitne door dikhte hain?
  mm
  Solution dikhao
  $1 / v = 1 / 12 + 1 / (- 10) = - 1 / 60$ , $v = - 60 c m$ ; $m = v / u = (- 60) / (- 10) = 6.0$ . Image $60 c m$ door hai, aur daante $12 m m$ door hain.
2. Magnifying power kya hai?
  guna
  Solution dikhao
  Lens se dekhne par image spacing $12 m m / 600 m m = 0.020 r a d$ angle banati hai. Nangi aankh se $25 c m$ par gear spacing $2.0 / 250 = 0.0080 r a d$ banati hai. Magnifying power $= 0.020 / 0.0080 = 2.5$ .
3. Answers alag kyun hain?
  Solution dikhao
  Image chhe guna badi hai, lekin woh door ( $60 c m$ ) bhi hai, isliye chhe guna badi nahi dikhti. Dikhne wala size aankh par bane angle se tay hota hai, aur magnifying power wahi compare karti hai. Dono tabhi barabar hote hain jab image near point par ho (Question 23).
23
NumericalMediumNCERT Exercises 9.23
Question 22 ka repairer chahta hai ki loupe gear ko jitna ho sake utna bada dikhaye aur image sharp rahe.
1. Gear lens se kitni door hona chahiye?
  cm
  Solution dikhao
  Image $v = - 25 c m$ par: $1 / u = 1 / v - 1 / f = - 1 / 25 - 1 / 12$ , isliye $u = - 8.1 c m$ .
2. Tab linear magnification kya hai?
  guna
  Solution dikhao
  $m = v / u = 25 / 8.11 = 3.08$ .
3. Kya ab yeh magnifying power ke barabar hai? Kyun?
  Solution dikhao
  Haan. Image $25 c m$ par hai, theek wahin jahan nangi aankh wala comparison hota hai, isliye angles compare karna sizes compare karne jaisa hi hai: dono $1 + D / f = 3.08$ ke barabar hain.
NumericalMediumNCERT Exercises 9.24
Usi loupe ( $f = 12 c m$ ) ke saath gear kahan rakhna chahiye taaki image mein uske daante $6.0 m m$ door dikhein? Kya repairer loupe ko bilkul aankh par rakh kar unhe sharp dekh payega?
cm
Solution dikhao
$m = 6.0 / 2.0 = 3$ , isliye $v = 3 u$ . Tab $1 / (3 u) - 1 / u = 1 / 12$ se $- 2 / (3 u) = 1 / 12$ milta hai, isliye $u = - 8.0 c m$ aur $v = - 24 c m$ .
Image $24 c m$ door hogi, $25 c m$ near point se theek andar, isliye zyadatar log use thoda dhundhla dekhenge ya thakaan mehsoos karenge.
25
Short answerMediumNCERT Exercises 9.25
Chhota sa answer do.
1. Hand lens ke theek peeche se dekhne par badi image aur khud object aankh par barabar angles banate hain. To lens asal mein tumhe kya fayda deta hai?
  Solution dikhao
  Lens ke bina object ko $25 c m$ se paas laakar sharp nahi dekha ja sakta. Lens tumhe use kuch centimetre door pakadne deta hai, jahan woh kaafi bada angle banata hai, jabki uski image aisi distance par rehti hai jahan aankh focus kar sake.
2. Tum hand lens ko stamp ke upar fixed pakadte ho aur apna sir usse peeche khinchte ho. Kya magnifying power same rehti hai?
  Solution dikhao
  Haan, ghatti hai. Aankh ke peeche jaane par image dwara aankh par bana angle kam hota hai, aur field ka kam hissa dikhta hai.
3. Jab $f$ chhota hone par $D / f$ badhta hi jaata hai, to dukaanein $100 \times$ hand lenses kyun nahi bechti?
  Solution dikhao
  Chhoti focal lengths ke liye chhote, bahut curved lenses chahiye. Unki spherical aur chromatic aberrations itni tezi se badhti hain ki image dhundhli aur rangeen ho jaati hai, isliye simple magnifiers lagbhag $10 \times$ par ruk jaate hain.
4. Microscope banane wala tube ke dono ends ke liye short-focus lenses chunta hai. Dono kyun, sirf ek kyun nahi?
  Solution dikhao
  $m \approx (L / f_{o}) (D / f_{e})$ : objective ka linear magnification aur eyepiece ka angular magnification dono unki focal lengths chhoti hone par badhte hain.
5. Microscope mein view tab sabse achha hota hai jab aankh eyepiece se chipkane ki jagah thodi peeche rakhi jaaye. Kyun, aur kitni peeche?
  Solution dikhao
  Objective se guzarne wali saari light eyepiece ke baad uske theek peeche ek chhote circle se converge hokar guzarti hai, jo eyepiece dwara bani objective ki image hai (eye ring). Pupil wahan ho to aankh saari light jama karti hai aur poora field dekhti hai. Aam instruments mein yeh eyepiece se $f_{e}$ se thoda zyada peeche, lagbhag ek-do centimetre hota hai.
26
NumericalHardNCERT Exercises 9.26
Ek kit mein ek objective lens ( $f = 1.0 c m$ ) aur ek eyepiece ( $f = 5.0 c m$ ) hai.
1. Inse ek microscope design karo jo final image near point par rakh kar $25$ guna magnify kare: object kahan hona chahiye, aur lenses kitne door hone chahiye?
  cm
  Solution dikhao
  $m_{e} = 1 + 25 / 5.0 = 6$ , isliye $∣ m_{o} ∣ = 25 / 6 = 4.17$ .
  $1 / v_{o} + 1 / ∣ u_{o} ∣ = 1 / f_{o}$ mein $v_{o} = 4.17 ∣ u_{o} ∣$ rakhne par: $(1 + 4.17) / (4.17 ∣ u_{o} ∣) = 1$ , isliye $∣ u_{o} ∣ = 1.24 c m$ aur $v_{o} = 5.17 c m$ .
  Eyepiece: $1 / u_{e} = - 1 / 25 - 1 / 5.0$ , isliye $∣ u_{e} ∣ = 4.17 c m$ . Lens separation $= 5.17 + 4.17 = 9.33 c m$ ; object objective se $1.24 c m$ door.
2. Lenses ko $9.33 c m$ door rakhte hue user relaxed aankh (final image infinity par) ke liye dobara focus karta hai. Ab magnifying power kya hai?
  guna
  Solution dikhao
  First image ab eyepiece ke focus par honi chahiye, objective se $9.33 - 5.0 = 4.33 c m$ : $1 / ∣ u_{o} ∣ = 1 - 1 / 4.33$ , $∣ u_{o} ∣ = 1.30 c m$ , $∣ m_{o} ∣ = 4.33 / 1.30 = 3.33$ .
  $m = 3.33 \times 25 / 5.0 = 16.7$ .
27
NumericalMediumNCERT Exercises 9.27
Ek hobby telescope mein $150 c m$ focal length ka objective aur $6.0 c m$ focal length ka eyepiece hai. Ise pehle relaxed aankh se use kiya jaata hai aur phir dobara focus kiya jaata hai taaki door ki pahadi ki final image $25 c m$ par ho.
1. Relaxed aankh ke saath magnifying power kya hai?
  guna
  Solution dikhao
  $m = 150 / 6.0 = 25$ .
2. Final image $25 c m$ par ho to yeh kya hai, aur kitne factor se badli hai?
  guna
  Solution dikhao
  $m = (f_{o} / f_{e}) (1 + f_{e} / D) = 25 \times (1 + 6.0 / 25) = 31$ , factor $1.24$ se badi.
28
NumericalMediumNCERT Exercises 9.28
Question 27 ka telescope $60 m$ oonchi paani ki tanki par point kiya jaata hai.
1. Relaxed aankh ke saath uske lenses kitne door hain?
  cm
  Solution dikhao
  $f_{o} + f_{e} = 150 + 6.0 = 156 c m$ .
2. Objective tanki ki $4.5 c m$ oonchi image banata hai. Tanki kitni door hai?
  km
  Solution dikhao
  Tanki $4.5 c m / 150 c m = 0.030 r a d$ angle banati hai, isliye uski distance $60 m / 0.030 = 2.0 \times 10^{3} m$ hai, yaani $2.0 k m$ .
3. Relaxed aankh ke saath tanki kitna angle bharti dikhti hai?
  °
  Solution dikhao
  $25 \times 0.030 r a d = 0.75 r a d$ , lagbhag $43^{\circ}$ (yahan small-angle approximation pehle hi kaafi khinch chuka hai).
NumericalHardNCERT Exercises 9.29
Ek chhote Cassegrain telescope mein $200 m m$ radius of curvature ka concave primary mirror aur $60 m m$ radius ka convex secondary mirror hai, jo primary ke $75 m m$ saamne rakha hai. Kisi taare ki final image kahan banti hai?
mm
Solution dikhao
Primary: $f_{1} = 100 m m$ , isliye reflected light uske $100 m m$ saamne ke point ki taraf converge hoti hai, jo secondary se $25 m m$ aage hai.
Secondary ke liye, ab uski taraf jaati light ke along naapne par: $u = + 25 m m$ (uske peeche virtual object) aur $f_{2} = + 30 m m$ (convex).
$1 / v = 1 / f_{2} - 1 / u = 1 / 30 - 1 / 25 = - 1 / 150$ , isliye $v = - 150 m m$ : secondary ke $150 m m$ saamne, wapas primary ki taraf, ek real image.
Yeh primary mirror ke $150 - 75 = 75 m m$ peeche hai, jahan light uske beech ke ched se bahar aati hai. Secondary $6$ guna magnify karta hai, isliye system chhoti tube mein band $600 m m$ focal length ki tarah kaam karta hai.
NumericalMediumNCERT Exercises 9.30
Ek sensitive galvanometer mein coil par laga chhota plane mirror light beam ko $1.2 m$ door ek scale par wapas reflect karta hai; current na ho to beam mirror par normally padti hai aur usi raaste wapas aati hai. Current behne par light spot scale par $8.4 c m$ khisakta hai. Coil kitne angle se ghoomi hai?
°
Solution dikhao
Reflected beam $\tan^{- 1} (0.084 / 1.2) = {4.0}^{\circ}$ ghoomi hai. Mirror, aur isliye coil, uska aadha ghooma: ${2.0}^{\circ}$ .
Isi doubling ki wajah se mirror aur lamba light lever chhote rotations ko padhna aasaan bana dete hain.
31
NumericalHardNCERT Exercises 9.31
Ek equiconvex glass lens ( $n = 1.50$ ) plane mirror par rakha hai. Uske upar axis ke along ek pin ko tab tak khiskaya jaata hai jab tak pin aur uski ulti image mil na jaayein; yeh lens se $24.0 c m$ upar hota hai.
1. Lens ke faces ka radius of curvature kya hai?
  cm
  Solution dikhao
  $f = 24.0 c m$ aur $1 / f = 0.50 \times 2 / R$ , isliye $R = 24.0 c m$ .
2. Lens aur mirror ke beech ek liquid ki kuch boondein daali jaati hain, aur milne ki distance $40.0 c m$ ho jaati hai. Liquid ka refractive index nikaalo.
  (koi unit nahi)
  Solution dikhao
  Combination ka $F = 40.0 c m$ hai, isliye liquid lens ka $1 / f_{l} = 1 / 40.0 - 1 / 24.0 = - 1 / 60$ , $f_{l} = - 60 c m$ .
  Plano-concave liquid lens ke liye $1 / f_{l} = - (n_{l} - 1) / R$ , isliye $n_{l} - 1 = 24.0 / 60 = 0.40$ aur $n_{l} = 1.40$ .
3. Agar liquid water ( $n = 1.33$ ) hota, to pin kahan milti?
  cm
  Solution dikhao
  $1 / f_{l} = - 0.33 / 24.0$ , $f_{l} = - 72.7 c m$ ; $1 / F = 1 / 24.0 - 1 / 72.7$ , isliye $F = 35.8 c m$ .
Quick answerMedium
Ek concave mirror screen par real object se teen guna size ki image banata hai. Agar object mirror se $20 c m$ door hai, to mirror ki focal length hai
$- 15 c m$ $- 30 c m$ $- 10 c m$ $+ 15 c m$
Answer dekho
Answer (A) $- 15 c m$
$u = - 20 c m$ , $v = - 60 c m$ : $1 / f = - 1 / 60 - 1 / 20 = - 1 / 15$ . $m = + 3$ (virtual) lene par $- 30 c m$ aata, lekin virtual image screen par nahi pakdi ja sakti.
Quick answerHard
Ek crown-glass lens ( $n = 1.52$ ) ki air mein focal length $20 c m$ hai. Water ( $n = 1.33$ ) mein uski power kitni hai, dioptres mein?
D
Answer dekho
Answer 1.37 D
$f_{w} = 20 \times 0.52 / (1.52 / 1.33 - 1) = 72.8 c m$ , isliye $P = 1 / 0.728 = 1.37 D$ .
34
Quick answerEasy
Compound microscope mein image banne ke stages ko object se shuru karke order mein lagao.
1. Object se light objective mein jaati hai
2. Woh image eyepiece ke focus par ya uske andar hoti hai
3. Eyepiece ek final, virtual, magnified image banata hai
4. Objective ek real, inverted, enlarged image banata hai
Answer dekho
Answer
Object se light objective mein jaati hai
Objective ek real, inverted, enlarged image banata hai
Woh image eyepiece ke focus par ya uske andar hoti hai
Eyepiece ek final, virtual, magnified image banata hai
Objective ki real image eyepiece ka object hai, jo magnifying glass ki tarah kaam karta hai.
35
Quick answerEasy
Har device ko us effect se match karo jis par woh depend karta hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
Endoscope : Fibres mein total internal reflection
Car ka rear-view mirror : Convex mirror ka chauda field
Shaving mirror : Concave mirror mein enlarged virtual image
n naapne wala spectrometer : Prism mein minimum deviation
Har device is chapter ka ek idea use karta hai.
Quick answerEasy
Light tabhi totally internally reflect ho sakti hai jab woh optically medium se optically medium ki taraf chal rahi ho.
Answer dekho
Answer (1) denser, (2) rarer
Refracted ray $90^{\circ}$ tak tabhi pahunch sakti hai jab woh normal se door mude; aisa denser se rarer medium mein jaate hue hota hai.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 9.1	Practice Q1
NCERT Exercises 9.2	Practice Q2
NCERT Exercises 9.3	Practice Q3
NCERT Exercises 9.4	Practice Q4
NCERT Exercises 9.5	Practice Q5
NCERT Exercises 9.6	Practice Q6
NCERT Exercises 9.7	Practice Q7
NCERT Exercises 9.8	Practice Q8
NCERT Exercises 9.9	Practice Q9
NCERT Exercises 9.10	Practice Q10
NCERT Exercises 9.11	Practice Q11
NCERT Exercises 9.12	Practice Q12
NCERT Exercises 9.13	Practice Q13
NCERT Exercises 9.14	Practice Q14
NCERT Exercises 9.15	Practice Q15
NCERT Exercises 9.16	Practice Q16
NCERT Exercises 9.17	Practice Q17
NCERT Exercises 9.18	Practice Q18
NCERT Exercises 9.19	Practice Q19
NCERT Exercises 9.20	Practice Q20
NCERT Exercises 9.21	Practice Q21
NCERT Exercises 9.22	Practice Q22
NCERT Exercises 9.23	Practice Q23
NCERT Exercises 9.24	Practice Q24
NCERT Exercises 9.25	Practice Q25
NCERT Exercises 9.26	Practice Q26
NCERT Exercises 9.27	Practice Q27
NCERT Exercises 9.28	Practice Q28
NCERT Exercises 9.29	Practice Q29
NCERT Exercises 9.30	Practice Q30
NCERT Exercises 9.31	Practice Q31

Quick check: MCQs

9.1 Introduction3 questions
9.2.3 Mirror equation4 questions
9.3 Refraction4 questions
9.4.1 Nature mein total internal reflection aur iske technological applications4 questions
9.5.4 Contact mein rakhe thin lenses ka combination4 questions
9.6 Prism se refraction4 questions
9.7.2 Telescope4 questions

Glossary · 15

Key terms (English + Hindi)

angle of minimum deviationन्यूनतम विचलन कोणnyuntam vichalan kon: Prism se ho sakne wala sabse kam deviation $D_{m}$ , jo tab milta hai jab ray symmetrically ( $i = e$ ) guzarti hai.Chapter mein pehli baar yahan aaya →

Cartesian sign conventionकार्तीय चिह्न परिपाटीkartiya chihn paripati: Distances pole ya optical centre se naapi jaati hain, incident light ke along positive; axis ke upar heights positive hain.Chapter mein pehli baar yahan aaya →
critical angleक्रांतिक कोणkrantik kon: Denser medium mein woh angle of incidence jis par refracted ray boundary ko chhoo kar chalti hai: $\sin i_{c} = n_{2} / n_{1}$ .Chapter mein pehli baar yahan aaya →

focal lengthफोकस दूरीphokas doori: Mirror ke pole ya lens ke optical centre se uske principal focus tak ki distance, $f$ .Chapter mein pehli baar yahan aaya →

magnificationआवर्धनaavardhan: Image height aur object height ka ratio, $m = h^{'} / h$ ; inverted image ke liye negative.Chapter mein pehli baar yahan aaya →
magnifying powerआवर्धन क्षमताaavardhan kshamata: Instrument ke saath aankh par bana angle kitne guna bada ho jaata hai: final image ka angle divided by seedhe dekhe gaye object ka angle (microscope ke liye near point par).Chapter mein pehli baar yahan aaya →

optical fibreप्रकाशिक तंतुprakashik tantu: Glass ya plastic ka patla dhaaga jiske core ka refractive index cladding se zyada hota hai, aur jo total internal reflection se light ko guide karta hai.Chapter mein pehli baar yahan aaya →

paraxial rayउपाक्षीय किरणupakshiya kiran: Principal axis ke paas ki ray jo usse chhota angle banati hai.Chapter mein pehli baar yahan aaya →
power of a lensलेंस की क्षमताlens ki kshamata: $f$ ko metres mein lekar $P = 1 / f$ , dioptres (D) mein naapi jaati hai; converging lenses ke liye positive.Chapter mein pehli baar yahan aaya →
principal focusमुख्य फोकसmukhya phokas: Woh point jahan principal axis ke parallel rays reflection ya refraction ke baad milti hain, ya jahan se diverge hoti lagti hain.Chapter mein pehli baar yahan aaya →

rayकिरणkiran: Ek line jiske along light energy chalti hai; rays ka bundle beam kehlata hai.Chapter mein pehli baar yahan aaya →
real imageवास्तविक प्रतिबिंबvastavik pratibimb: Aisi image jahan object point se aayi rays sach mein milti hain; ise screen par pakda ja sakta hai.Chapter mein pehli baar yahan aaya →
refractive indexअपवर्तनांकapvartanank: Kisi medium ke liye $n = c / v$ ; pair ke liye $n_{21} = n_{2} / n_{1} = \sin i / \sin r$ .Chapter mein pehli baar yahan aaya →

total internal reflectionपूर्ण आंतरिक परावर्तनpoorn aantarik paravartan: Denser medium ke andar light ka poora reflection, jab woh critical angle se zyada angle par rarer medium se milti hai.Chapter mein pehli baar yahan aaya →

virtual imageआभासी प्रतिबिंबaabhasi pratibimb: Aisi image jahan se rays sirf diverge hoti lagti hain; ise screen par nahi pakda ja sakta.Chapter mein pehli baar yahan aaya →

9.1 Introduction

9.2 Spherical mirrors se light ka reflection

9.2.1 Sign convention

9.2.2 Spherical mirrors ki focal length

9.2.3 Mirror equation

9.3 Refraction

9.4 Total internal reflection

9.4.1 Nature mein total internal reflection aur iske technological applications

9.5 Spherical surfaces par aur lenses se refraction

9.5.1 Spherical surface par refraction

9.5.2 Lens se refraction

9.5.3 Lens ki power

9.5.4 Contact mein rakhe thin lenses ka combination

9.6 Prism se refraction

9.7 Optical instruments

9.7.1 Microscope

9.7.2 Telescope

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye