Wave Optics: Class 12 Physics Chapter 10 Hinglish mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

95 min
15 practice questions
26 MCQs

Is chapter mein aap kya seekhenge

Samjho ki wavefront kya hota hai, aur Huygens' construction se spherical aur plane wavefronts ki aage ki positions nikaalo
Huygens' principle se refraction aur reflection ke laws derive karo, aur samjhao ki boundary par speed aur wavelength kyun badalte hain jabki frequency nahi badalti
Critical angle ko refractive indices se jodo, aur batao ki prisms, lenses aur mirrors wavefronts ki shape kaise badalte hain
Coherent aur incoherent sources mein farq karo, aur path aur phase difference se interference predict karo, I = 4I0 cos²(φ/2) samet
Young's double-slit experiment samjhao aur fringe positions aur fringe width calculate karo
Single-slit diffraction describe karo, uske minima dhoondho, aur samjhao ki yeh asli double-slit pattern ko kaise shape deta hai
Polarisation ko transverse waves ki property ke roop mein samjhao, polaroids kaise kaam karte hain yeh batao aur Malus' law apply karo

NCERT Physics Class 12, Chapter 10, pp. 255–273 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

10.1 Introduction

Dhoop mein ek soap bubble ko dekho, ya geeli sadak par phaile oil ki chamak ko, to tumhe rang ki aisi bands dikhti hain jo kisi prism ne nahi banayi. Tube light ke saamne do ungliyan lagbhag milakar pakdo aur gap mein se aankhein sikod kar dekho: usmein patli dark lines dikhti hain. Ray optics, jo light ko seedhi lines maanta hai, in dono mein se kisi ko explain nahi karta. Is chapter mein hum light ko ek wave maante hain, aur yeh ek idea reflection aur refraction ke saath-saath in naye effects ko bhi samjha deta hai.

Light particles ki stream hai ya wave? Dono pictures explain karti hain ki ray barabar angles par kyun reflect hoti hai aur water surface par kyun mudti hai, isliye sau saal se bhi zyada tak koi bhi side doosri ko galat saabit nahi kar paayi. Fig. 10.1 woh ek jagah dikhata hai jahan dono ke raaste alag ho jaate hain: water ke andar light kitni tez chalti hai.

4/4Foucault ka 1850 ka measurement: water mein light dheere chalti hai, jaisa wave model kehta hai.

Fig. 10.1Bending same, predictions ulti. Measurement ne faisla waves ke favour mein kiya.NCERT 10.1 · p. 255

Light ki ek ray air (upar) se water (neeche) mein jaati hai aur dashed normal ki taraf mudti hai; angles i aur r marked hain. Step 1 sirf mudi hui ray dikhata hai. Step 2 right mein particle model jodta hai: dashed components wale do velocity arrows, jinmein water wale ka sideways part same hai lekin downward part lamba hai, isliye model predict karta hai ki water mein light tez hai. Step 3 iski jagah ray ke aar-paar chhote blue wavefronts dikhata hai, jo water mein zyada paas-paas hain, aur label kehta hai ki wave model dheemi light predict karta hai. Step 4 wavefronts rakhta hai aur measured speeds ka ek box jodta hai: air mein lagbhag 3.00 times ten to the 8 metres per second aur water mein 2.25 times ten to the 8, isliye wave model jeet jaata hai.

Particle picture mein, water tak pahunchne wale light ke particle ko surface ki taraf ek kheench lagti hai. Yeh kheench uski velocity ke surface ke perpendicular wale part ko badha deti hai aur surface ke along wale part ko waisa hi chhod deti hai, isliye particle normal ki taraf mudta hai aur tez bhi ho jaata hai. Wave picture mein, wavefront ka jo end pehle water mein ghusta hai woh slow ho jaata hai, baaki front abhi bhi tez chal raha hota hai, aur isliye poora front normal ki taraf ghoom jaata hai. Bending same hai; speed ke baare mein predictions ulti hain.

Kuch metres water mein light ka time naapna 1850 tak namumkin tha, jab Léon Foucault ne ek beam ko second mein saikdon baar ghoomte mirror se bounce karaya, ek baar air se hokar aur ek baar water ki tube se hokar. Water wale path ke liye lautne wala spot zyada khiska, matlab wahan light ne zyada time liya. Water mein light dheemi hai: air mein $3.00 \times 10^{8} m / s$ ke against lagbhag $2.25 \times 10^{8} m / s$ . Wave model ek aisa test paas kar gaya jismein particle model fail ho gaya.

Timeline

Particles ya waves?

1637
Descartes
Light ko particles maanta hai aur usi picture se refraction ka law derive karta hai.
1678
Huygens
Ek wave theory aur secondary wavelets wala construction propose karta hai (1690 mein chhapa).
1704
Newton ki Opticks
Newton ki saakh ki wajah se particle (corpuscular) model sau saal tak maana hua view bana rehta hai.
1801
Young
Two-slit interference: ek source ki light ko do mein baanto, to woh bright aur dark bands deti hai.
1818
Fresnel
Diffraction ki ek wave theory jo edges aur slits par bane patterns ko detail mein predict karti hai.
1850
Foucault
Water mein light air se dheemi hai, jaisa wave model predict karta hai.
1865
Maxwell
Light ek electromagnetic wave hai aur use kisi material medium ki zaroorat nahi.
1887
Hertz
Radio wavelengths par electromagnetic waves banata aur detect karta hai.
1895
J. C. Bose
Calcutta mein millimetre-length electromagnetic waves par kaam karte hain.

Tab tak wave model ke paas aur saboot bhi the. 1801 se Thomas Young ne dikhaya ki ek source ki light ko do beams mein baantkar phir overlap karaya jaaye, to woh bright aur dark bands deti hai, jo waves ki pehchaan hai (Section 10.5). Aise experiments ne light ki wavelength bhi naapi, aur woh bahut chhoti nikli: visible light lagbhag $400$ se $700 n m$ tak hoti hai, yaani insaan ke ek baal ki chaudaai mein green light ki lagbhag 150 wavelengths aa jaati hain. Kuch centimetres chaude mirror ya lens ke saamne itni chhoti wave lagbhag bilkul seedhi rays ke bundle jaisa behave karti hai. Jab wavelength ko na ke barabar chhota maan liya jaaye, to wave optics hi ray optics ban jaata hai, aur tab ray bas woh direction hai jismein wave apni energy le jaati hai.

Ek objection bacha tha. Us samay ki har jaani-maani wave, ripples se lekar sound tak, kisi material se hokar chalti thi, to light Sun aur Earth ke beech ki khaali space kaise paar kar sakti thi? 1860s mein Maxwell's equations ne iska jawab diya: light ek electromagnetic wave hai, electric aur magnetic fields ka ek pattern jo ek doosre ko banaye rakhte hain aur jinhe kisi medium ki zaroorat nahi (Part I, Chapter 8). Hertz ne 1887 mein radio wavelengths par aisi waves banayin, aur J. C. Bose aur G. Marconi ne jald hi unka practical use kiya.

Chaar sawaal baaki chapter ko organise karte hain. Wave model reflection aur refraction ke laws kaise dobara deta hai (Section 10.2, Section 10.3)? Jahan do waves overlap karti hain wahan kya hota hai, general mein (Section 10.4) aur Young's experiment mein (Section 10.5)? Jab wave kisi patli opening se nikalti hai to kya karti hai (Section 10.6)? Aur hum kaise jaanein ki light waves transverse hain (Section 10.7)?

Quick check

3 mein se 1

Air se water mein jaate hue light normal ki taraf mudti hai. Particle model ne water mein uski speed ke baare mein kya predict kiya?

Air se zyadaAir se kamAir wali speed ke barabar

Answer dekho

Answer (A) Air se zyada

Particle model mein surface light ko andar kheenchti hai, jisse normal ke along velocity component badhta hai, isliye water mein light ko tez hona chahiye. Wave model dheemi predict karta hai, aur Foucault ne dheemi hi naapi.

10.2 Huygens principle

Socho ek bade khaali hall ke beech mein ek chhota loudspeaker steady note baja raha hai. Kisi ek instant par, har us point ko mark karo jahan air pressure abhi-abhi apne peak par pahuncha hai. Yeh points speaker ke around ek sphere par hain: sab ek hi distance par hain, isliye disturbance un tak ek saath pahuncha aur woh in step oscillate karte hain.

Aisi surface, jis par har point apne oscillation ki same stage (phase) par ho, wavefront (तरंगाग्र) kehlaati hai. Time ke saath wavefront wave speed $v$ se bahar ki taraf badhta hai, aur wave apni energy uske right angle par le jaati hai. Isiliye rays, yaani woh lines jinke along energy behti hai, kisi bhi aise medium mein wavefronts ke perpendicular banayi jaati hain jismein light ki speed har direction mein same ho.

Definition

Wavefront

Aisi surface jis par wave ka har point same phase mein oscillate karta hai (constant phase ki surface). Yeh wave speed se chalti hai, aur rays iske perpendicular hoti hain.

ExampleChhote lamp ke around wavefronts lamp par centred spheres hote hain.

Wavefront ki shape source par aur is baat par depend karti hai ki tum kitni door ho (Fig. 10.2). Point source spherical wavefronts bhejta hai, aur rays radially phailti hain. Source se door, jo bhi patch tum dekho woh ek bahut bade sphere ka chhota sa tukda hai aur lagbhag flat hai: ek plane wavefront, parallel rays ke saath. Tumhari khidki par aati sunlight iska example hai. Sun $1.5 \times 10^{11} m$ door hai, isliye $1 m$ chaudi khidki ke across uska wavefront sirf lagbhag $8 \times 10^{- 13} m$ ubhra hota hai, yaani wavelength ka lagbhag ek millionth hissa. Usi khidki se $2 m$ door rakhe bulb ke liye yeh ubhaar lagbhag $6 c m$ hai, isliye uske wavefronts saaf curved hain.

Fig. 10.2Point source ke paas wavefronts spheres hote hain aur rays phailti hain; door jaakar wavefront ka chhota patch lagbhag plane hota hai aur rays parallel hoti hain.NCERT Fig. 10.1 · p. 257

Dashed line se alag kiye do panels. Left panel, source ke paas: ek chhota yellow source, uske around ripples jaise paanch blue circular arcs, aur source se radially phailte paanch orange arrows (rays). Right panel, bahut door: paanch blue lines jo lagbhag seedhi aur parallel hain, bas zara si curved, aur unke aar-paar parallel chalti teen orange rays; label kehta hai ki yeh lagbhag planes hain.

Huygens ki samajh ek recipe thi, yeh nikaalne ki ki agar hum jaante hain ki wavefront abhi kahan hai to thodi der baad woh kahan hoga: present wavefront ke har point ko apne aap mein ek chhota source maan lo.

PrincipleHuygens' principle

Wavefront ka har point secondary wavelets ka source banta hai, jo wave ki speed se phailti hain. Time $τ$ ke baad naya wavefront woh surface hai jo aage ki side par in sab wavelets ko chhooti hai: unka forward envelope.

Fig. 10.3 is recipe ko ek spherical wavefront ke hisse par apply karta hai. Har marked point ke around radius $v τ$ ka ek sphere banao. Jo surface bahar ki taraf in sabko chhooti hai woh phir se source par centred ek sphere hai, jo $v τ$ se bada hai: wave bas bahar ki taraf badh gayi hai. Plane wavefront ke liye yahi construction $v τ$ aage ek parallel plane deta hai. Dono cases mein rays, jo wavefronts ke normals hain, seedhi lines mein chalti rehti hain, jaisa uniform medium mein hona chahiye.

3/4Aage ki taraf saare wavelets ko chhoone wali surface time $τ$ par wavefront hai.

Fig. 10.3Wavefront ke har point se wavelets, aur unka envelope jo naya wavefront banta hai.NCERT Fig. 10.2 · p. 257

Bottom centre mein ek yellow source hai. Step 1: uske upar ek blue arc time zero par wavefront hai, jis par paanch dots marked hain. Step 2: har dot ke around radius v tau ka ek grey circle banaya gaya hai, jis par wavelets likha hai. Step 3: bahar se saare circles ko chhoota ek mota orange arc dikhta hai, yaani time tau par wavefront, aur dots se ustak bahar jaate paanch black arrows (rays). Step 4: andar ki taraf, source ke paas, circles ko chhoota ek dashed vermilion arc joda jaata hai, jis par likha hai ki koi backwave nahi.

Quick check

3 mein se 1

Ek lambi, seedhi tube light apni poori length mein barabar chamakti hai. Uske paas, uske wavefronts ki shape kya hai?

Tube ke beech par centred spheresTube ke parallel planesCylinders jinka axis tube ke along hai

Answer dekho

Answer (C) Cylinders jinka axis tube ke along hai

Tube ki line se same distance wale har point tak light ek hi time par pahunchti hai, isliye constant phase ki surfaces uske around cylinders hain. Spheres ke liye point source chahiye, aur planes ke liye bahut door ka source.

10.3 Huygens principle se plane waves ka refraction aur reflection

Huygens' principle ka asli fayda tab dikhta hai jab wavefront do media ki boundary se milta hai. Front ka jo hissa boundary paar kar chuka hai woh us hisse se alag speed se chalta hai jo abhi paar nahi hua, aur construction bilkul dikhata hai ki front kaise mudta hai.

10.3.1 Plane wave ka refraction

Maan lo light medium 1 mein speed $v_{1}$ se aur medium 2 mein $v_{2}$ se chalti hai, jahan $v_{2} < v_{1}$ (jaise air se glass mein). Ek plane wavefront flat boundary se ek angle par milta hai, isliye uska ek end doosre se pehle boundary ko chhoota hai (Fig. 10.4).

DerivationWavelets se Snell's law

Wavefront ke jis end ne abhi-abhi boundary chhooi hai use $P$ kaho, aur doosre end ko $Q$ . $Q$ se light ko boundary par $R$ tak pahunchne ke liye medium 1 mein abhi distance $Q R$ chalna hai. Maan lo ismein time $τ$ lagta hai, to $Q R = v_{1} τ$ . Isi time mein, $P$ se shuru hui wavelet medium 2 mein radius $v_{2} τ$ tak badhti hai.

Huygens' principle ke hisaab se, jab $Q$ se light $R$ par pahunchti hai, to refracted wavefront $R$ se guzarne wala woh plane hai jo is wavelet ko chhoota hai. Contact point ko $S$ kaho. Tab $P S = v_{2} τ$ , aur $P S$ , $R S$ ke perpendicular hai, kyunki radius tangent se right angle par milta hai.

Wavefront aur boundary ke beech ka angle ray aur normal ke beech ke angle ke barabar hota hai, kyunki doosri pair ki har line pehli pair ki apni partner line se ek right angle ghoomi hui hai. Isliye angle $Q P R$ angle of incidence $i$ hai, aur angle $P R S$ angle of refraction $r$ hai. Right-angled triangles $P Q R$ aur $R S P$ ka hypotenuse $P R$ common hai, isliye

\sin i = \frac{Q R}{P R} = \frac{v_{1} τ}{P R}

(10.1)

\sin r = \frac{P S}{P R} = \frac{v_{2} τ}{P R}

(10.2)

Ek ko doosre se divide karne par unknowns $τ$ aur $P R$ cancel ho jaate hain:

\frac{\sin i}{\sin r} = \frac{v_{1}}{v_{2}}

(10.3)

4/4Common side $P R$ wale right-angled triangles se (10.1) aur (10.2) milte hain.

Fig. 10.4Jitne time mein

Q

se light

R

tak pahunchti hai, utne mein

P

se wavelet dheemi glass mein sirf

v_{2} τ

tak badhti hai, isliye front mud jaata hai.NCERT Fig. 10.4 · p. 258

Boundary ek vertical line hai, left mein air aur right mein shaded glass. Step 1: upar-left se teen parallel orange rays aati hain; ek mota blue wavefront PQ unke perpendicular hai, jismein P boundary par hai aur Q peeche; ek dashed blue arrow path QR dikhata hai, jiski length v1 tau hai, jo Q se light ko boundary par R tak pahunchne ke liye abhi tay karna hai. Step 2: glass mein P ke around radius v2 tau ka ek dashed orange semicircle banaya gaya hai, jo QR se chhota hai. Step 3: R se semicircle ko S par chhoone wali ek moti orange line refracted wavefront hai; glass mein rays normal ki taraf mudkar aage badhti hain, aur P par angle i aur R par angle r marked hain. Step 4: triangles PQR aur RSP shaded hain aur ratios sin i = v1 tau over PR aur sin r = v2 tau over PR dikhaye gaye hain.

Equation (10.3) refraction ka law hai, aur ismein ek prediction chhupi hai. Normal ki taraf mudne wali ray ke liye $r < i$ , isliye $\sin i / \sin r > 1$ aur isliye $v_{2} < v_{1}$ : jis medium mein light normal ki taraf mudti hai, usmein use dheere chalna hi hoga. Yahi Fig. 10.1 mein wave model wali side hai, aur Foucault ka measurement isse match karta hai.

Kisi medium ka refractive index vacuum mein light ki speed $c$ ko medium mein uski speed se compare karta hai:

n_{1} = \frac{c}{v_{1}}

(10.4)

n_{2} = \frac{c}{v_{2}}

(10.5)

(10.3) mein $v_{1} = c / n_{1}$ aur $v_{2} = c / n_{2}$ rakhne par Snell's law apne jaane-pehchaane form mein milta hai:

Snell's law

n_{1} \sin i = n_{2} \sin r

(10.6)

Wavelength ka kya hota hai? $τ$ ko wave ka theek ek period $T$ chuno. Ek period mein wave ka har hissa ek wavelength aage badhta hai, isliye $Q R = λ_{1}$ aur $P S = λ_{2}$ , aur (10.1) aur (10.2) se milta hai

\frac{λ_{1}}{λ_{2}} = \frac{v_{1}}{v_{2}} ya \frac{v_{1}}{λ_{1}} = \frac{v_{2}}{λ_{2}}

(10.7)

Kyunki $v / λ$ frequency $ν$ hai, isliye boundary ke dono taraf frequency same rehti hai. Sirf speed aur wavelength badalte hain, aur dono same factor $n_{1} / n_{2}$ se badalte hain. Air se refractive index $1.5$ wale glass mein ghusne wali light ki frequency wahi rehti hai, lekin uski speed aur wavelength air wali values ki do-tihaai reh jaati hain.

Example

Glass mein ghusti laser beam

Ek green laser pointer ( $λ = 532 n m$ air mein) refractive index $1.52$ wale glass block par $50^{\circ}$ ke angle of incidence par chamakta hai. Nikaalo (a) angle of refraction, (b) glass mein light ki speed, aur (c) glass mein light ki frequency aur wavelength. $c = 3.00 \times 10^{8} m / s$ lo aur air ko vacuum maano.

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

History seNCERT Christiaan Huygens · p. 259

Christiaan Huygens (1629–1695)

Huygens The Hague mein bade hue, ek diplomat aur kavi ke bete, aur solah saal ki umar mein university jaane tak ghar par hi padhe. Ek telescope se, jiske lenses unhone aur unke bhai ne khud ghise the, unhone 1655 mein Saturn ka sabse bada moon, Titan, discover kiya, aur samjha ki planet ke ajeeb 'handles' asal mein ek patli, flat ring hain jo use kahin nahi chhooti. 1656 mein unhone pehli pendulum clock banayi, jisne achhi clocks ki galti ko din mein lagbhag ek chauthai ghante se ghatakar ek minute se bhi kaafi kam kar diya. Unki Treatise on Light, jo 1678 mein likhi gayi aur 1690 mein chhapi, ne woh wavelet construction diya jo tumne abhi use kiya, aur Iceland spar crystals mein dikhne wali double images ko explain kiya. Unhone probability par pehli kitaabon mein se ek bhi likhi.

10.3.2 Rarer medium par refraction

Ab situation ulti karo, taaki light glass ya water se air mein jaaye aur $v_{2} > v_{1}$ ho. Construction wahi hai, lekin ab medium 2 mein wavelet us speed se tez badhti hai jisse $Q$ ki light boundary tak ka gap kam karti hai. Refracted wavefront doosri taraf jhukta hai aur ray normal se door mudti hai, $r > i$ , aur Snell's law ab bhi lagta hai (Fig. 10.5).

Jaise-jaise $i$ badhta hai, $r$ pehle $90^{\circ}$ tak pahunchta hai. Jis angle of incidence par aisa hota hai, use critical angle $i_{c}$ (क्रांतिक कोण) kehte hain. (10.6) mein $r = 90^{\circ}$ rakhne par milta hai

Critical angle

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

(10.8)

Construction yeh bhi dikhata hai ki is angle ke aage kuch bhi refract kyun nahi ho sakta. (10.2) ke hisaab se, $i = i_{c}$ par $P$ se wavelet time $τ$ mein bas itni badhti hai ki khud $R$ tak pahunch jaaye: $v_{2} τ = P R$ , aur refracted wavefront boundary ke perpendicular khada hota hai, uske saath-saath sarakta hua. $i > i_{c}$ ke liye, jab tak $Q$ se light wahan pahunchti hai, wavelet $R$ ko paar kar chuki hoti hai, aur sphere ke andar ke kisi point se guzarne wala koi bhi plane sphere ko chhoo nahi sakta. Isliye koi refracted wave hoti hi nahi, aur saari light denser medium mein wapas reflect ho jaati hai. Yahi total internal reflection hai jo tumne pichhle chapter mein padha.

2/3 $i = i_{c} = {41.8}^{\circ}$ : refracted ray surface ke saath-saath sarakti hai ( $r = 90^{\circ}$ ).

Fig. 10.5Glass se air mein: normal se door mudna, critical angle par surface ke saath sarakna, phir total internal reflection. Index

1.5

wale glass ke liye banaya gaya, isliye

i_{c} = {41.8}^{\circ}

.NCERT Fig. 10.5 · p. 260

Picture ke neeche wale hisse mein glass hai aur upar wale mein air, aur jahan ray surface se milti hai wahan ek dashed normal hai. Step 1: glass ke andar ek ray chhote angle of incidence i par surface se milti hai aur bade angle r par air mein nikalti hai, normal se door mudkar, saath mein ek halki dashed reflected ray. Step 2: ray critical angle par surface se milti hai, aur refracted ray 90 degrees par surface ke saath-saath sarakti hai; label kehta hai sine of i c equals n2 over n1. Step 3: angle of incidence aur bhi bada hai; air mein koi ray nahi, aur ek strong reflected ray wapas neeche glass mein jaati hai, jis par total internal reflection likha hai.

Example

Diamond kyun chamakta hai

Diamond ka refractive index $2.42$ hai. (a) Diamond se air mein nikalne ki koshish karti light ke liye critical angle nikaalo. (b) Ek cut diamond ke andar light ek face se $35^{\circ}$ ke angle of incidence par milti hai. Uske saath kya hota hai? (c) Kya yahi light index $1.50$ wale glass se bahar nikal jaati?

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

10.3.3 Plane surface se plane wave ka reflection

DerivationWavelets se reflection ka law

Maan lo ek plane wavefront $P Q$ ek plane mirror se angle $i$ par milta hai (Fig. 10.6). End $P$ mirror ko pehle chhoota hai. $Q$ se light ko mirror par $R$ tak pahunchne mein time $τ$ lagta hai, isliye $Q R = v τ$ . Isi beech $P$ se wavelet usi medium mein wapas phailti hai aur usi radius $v τ$ tak pahunchti hai.

Reflected wavefront $R$ se guzarne wala woh plane hai jo is wavelet ko chhoota hai, maan lo $S$ par. Triangles $P Q R$ aur $R S P$ dono mein ek right angle hai ( $Q$ par aur $S$ par), dono ka hypotenuse $P R$ common hai, aur $Q R = P S = v τ$ . Isliye woh congruent hain, to angle $Q P R$ angle $S R P$ ke barabar hai: incident aur reflected wavefronts mirror ke saath barabar angles banate hain. Yeh angles un angles ke barabar hain jo rays normal ke saath banati hain, isliye $i = r$ , yahi reflection ka law hai.

3/3Tangent $R S$ reflected wavefront hai; congruent triangles se $i = r$ milta hai.

Fig. 10.6

P

se wavelet usi radius

v τ

tak badhti hai jitna

Q

se light chalti hai, isliye dono shaded triangles congruent hain aur

i = r

.NCERT Fig. 10.6 · p. 261

Picture ke beech mein ek vertical mirror khada hai, jiski right side hatched hai. Step 1: upar-left se mirror ki taraf teen parallel orange rays aati hain; ek mota blue wavefront PQ unhe kaat-ta hai, jismein P mirror par hai aur Q peeche, aur Q se mirror par R tak ek dashed blue arrow woh distance v tau dikhata hai jo Q se light ko abhi chalna hai. Step 2: mirror ki saamne wali side par P ke around radius v tau ka ek dashed orange arc banaya gaya hai. Step 3: R se arc ko S par chhoone wali moti orange line reflected wavefront hai; reflected rays wapas neeche left ki taraf jaati hain, triangles PQR aur RSP shaded hain, barabar angles i aur r marked hain, aur ek note kehta hai ki triangles congruent hain, isliye i equals r.

Reflection aur refraction samajh aane ke baad, hum kisi bhi optical device se guzarte wavefront ko follow kar sakte hain. Ek hi rule sab par lagta hai: front ka jo hissa slow glass mein zyada time bitaata hai, ya mirror se wapas aane se pehle jise zyada chalna padta hai, woh baaki se peeche reh jaata hai, aur front ki shape usi hisaab se badal jaati hai (Fig. 10.7, Table 10.1). Kyunki rays front ke perpendicular hoti hain, uski nayi shape turant bata deti hai ki rays kahan jaati hain.

Table 10.1Optical devices plane wavefront ke saath kya karte hain

	Front ka sabse zyada roka gaya hissa	Nikalne wala front	Rays
Thin prism	mote base se guzarne wala hissa	ek tircha plane	base ki taraf mudti hain
Convex lens	centre (sabse mota glass)	focus ki taraf simatta sphere	converge karti hain
Concave lens	edges (sabse mota glass)	lens ke peeche ke ek point se phailta sphere	diverge karti hain
Concave mirror	centre (edges pehle reflect hote hain)	focus ki taraf simatta sphere	converge karti hain
Convex mirror	edges (centre pehle reflect hota hai)	mirror ke peeche ke ek point se phailta sphere	diverge karti hain

2/4Mota centre beech ke hisse ko late karta hai: front focus $F$ par converge hota hai.

Fig. 10.7Jahan bhi wave glass mein zyada time bitaati hai, ya mirror tak jaldi pahunchti hai, front ke us hisse ki timing badal jaati hai, aur front ki shape badal jaati hai.NCERT Fig. 10.7 · p. 261

Har step mein left se plane blue wavefronts aate hain. Step 1, prism: ek triangular glass prism jiska base neeche hai; usse nikalne wale orange wavefronts tirche hain, upar wala hissa aage, aur ek arrow dikhata hai ki light neeche base ki taraf mud gayi. Step 2, convex lens: lens ke baad orange arcs right ki taraf khokhle hain aur focus F par simat jaate hain. Step 3, concave lens: lens ke baad orange arcs right ki taraf ubhre hain aur badhte jaate hain, unka centre lens ke saamne ek khokhla point F hai. Step 4, concave mirror: right mein ek curved mirror orange arcs wapas bhejta hai jo apne saamne ke focus F ki taraf khokhle hain, aur ek arrow dikhata hai ki reflected light wapas left ki taraf ja rahi hai.

Example 10.1NCERT Example 10.1 · p. 262

Boundary par frequency, speed aur energy

Ek green laser beam ( $532 n m$ air mein) swimming pool par daali jaati hai. Paani ke neeche ek swimmer ko wahan green spot dikhta hai jahan beam andar aati hai, aur beam ka kuch hissa ek darshak ki taraf reflect hota hai.

(a) Swimmer aur darshak dono ko green dikhta hai. Water ke andar aur reflected beam mein light ki frequency kaun tay karta hai, aur yeh incident light ki frequency ke barabar kyun hai? (b) Water ke andar beam lagbhag $2.25 \times 10^{8} m / s$ se chalti hai, phir bhi glass ki side wall se pool ke bahar nikalne ke baad woh phir se $3.00 \times 10^{8} m / s$ par aa jaati hai. Use phir se tez karne ki energy kahan se aati hai? (c) Same colour ka ek doosra laser das guna bright hai. Uske ek photon ki energy, aur woh har second jitne photons deta hai, unhe pehle laser se compare karo.

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

(a) Surface koi frequency nahi chunti; water ke atoms aane wali wave ko respond karte hain. Wave ka oscillating electric field har atom ke electrons ko wave ki apni frequency par forced oscillation mein daal deta hai, aur yeh oscillating charges usi frequency par radiate karte hain. Reflected aur refracted waves isi re-radiated light se bani hain, isliye unhe incident frequency hi milti hai. (Surface par crests na jama ho sakte hain na gaayab, isliye har second ek taraf se jitne crests paar karte hain utne hi doosri taraf.)
(b) Koi energy nahi chahiye, kyunki slow hone mein koi energy gayi hi nahi thi. Light ki speed us medium se tay hoti hai jismein woh hai, jabki beam jo energy le jaati hai woh uske amplitude par (aur photon-dar-photon, uski frequency par) depend karti hai. Water mein slow hona aur air mein phir tez hona energy ke len-den nahi hain. Beam har surface par thodi energy khoti zaroor hai, lekin sirf isliye ki wahan kuch light reflect ho jaati hai.

Quick check

4 mein se 1

Air mein

600 n m

wavelength wali orange light refractive index

1.50

wale glass mein jaati hai. Glass mein uski wavelength kya hai, nanometres mein?

Answer dekho

Answer 400 nm

λ = λ_{0} / n = 600 / 1.50 = 400 n m

. Frequency nahi badalti, isliye light ab bhi orange hai.

10.4 Waves ka coherent aur incoherent addition

Jab do waves same point par pahunchti hain to kya hota hai? Ripples, sound aur light, sabke liye experiment ek seedha jawab deta hai, superposition principle (अध्यारोपण का सिद्धांत):

PrincipleSuperposition ka principle

Jahan waves overlap karti hain, wahan har wave medium ko bilkul waise hi disturb karti hai jaise baaki waves ke na hone par karti, aur kisi bhi point aur instant par asli disturbance in alag-alag disturbances ka vector sum hota hai.

Light ke liye 'displacement' wave ka electric field hai. Interference yahi principle hai jo same frequency ki do waves par lagaya jaata hai, aur humein kuch interesting dikhega ya nahi, yeh is par depend karta hai ki dono waves step mein rehti hain ya nahi.

Ek vibrating arm par do prongs lagao taaki woh ek saath ripple tank mein dubki lagayein. Har prong circular ripples ka source hai, aur kyunki ek hi motor dono ko chalata hai, woh hamesha step mein upar-neeche hote hain. Jin sources ka phase difference is tarah constant rehta hai, unhe coherent sources (कला-संबद्ध स्रोत) kehte hain. Fig. 10.8 unki ripples ka ek snapshot dikhata hai: crests solid circles ki tarah aur troughs dashed circles ki tarah.

Kuch lines ke along crest hamesha crest se aur trough trough se milta hai, aur paani ek source ke comparison mein do guna upar uthta hai. Doosri lines ke along crest hamesha trough se milta hai, aur paani mushkil se hilta hai. Kisi point $P$ par kya hoga, yeh sirf path difference $Δ = S_{2} P - S_{1} P$ (पथांतर) par depend karta hai, jise wavelengths mein naapa jaata hai.

Agar $Δ = 0$ , jaise $S_{1} S_{2}$ ke perpendicular bisector par kahin bhi, to dono waves step mein nikalti hain aur step mein pahunchti hain. Agar $Δ = λ$ ya $3 λ$ , to door wale source ki wave ek ya teen poore cycles late pahunchti hai, jisse woh phir step mein aa jaati hai. Lekin agar $Δ = λ / 2$ ya $3.5 λ$ , to late wave odd number of half cycles peeche hai, isliye crest trough ke saath pahunchta hai. Path difference $Δ$ ek phase difference $ϕ = 2 π Δ / λ$ ke barabar hai. $n = 0, 1, 2, \dots$ ke saath, dono conditions hain

Constructive interference

∣ S_{2} P - S_{1} P ∣ = n λ

(10.9)

Destructive interference

∣ S_{2} P - S_{1} P ∣ = (n + \frac{1}{2}) λ

(10.10)

Har condition ek curve chunti hai. Do fixed points se jin points ki distances ka difference fixed ho, woh ek hyperbola par hote hain, isliye Fig. 10.8 mein strong motion ki lines (antinodal lines) aur stillness ki lines (nodal lines) hyperbolas hain. Yeh sources ke beech se pankhe ki tarah phailti hain aur door jaakar lagbhag seedhi ho jaati hain, aur unke aar-paar rakhi screen strong aur weak spots ki ek row dikhati hai.

2/3Green: $Δ = n λ$ , crest crest se milta hai. Dashed: $Δ = (n + \frac{1}{2}) λ$ , crest trough se milta hai.

Fig. 10.8Step mein chalaye gaye do sources. Jahan path difference wavelengths ki whole number ho wahan strong motion, aur jahan half wavelengths ki odd number ho wahan stillness. Yahan

S_{1} S_{2} = 5 λ

.NCERT Fig. 10.8 · p. 262

Left mein do yellow sources S1 aur S2 ek ke upar ek hain. Step 1: har ek ke around solid blue circles crests aur dashed grey circles troughs dikhate hain, jo right ki taraf overlap karte hain. Step 2: circles halke pad jaate hain aur sources ke beech se mote green curves (antinodal lines, jahan crest crest se milta hai) pankhe ki tarah phailte hain, jin par path difference 0 aur lambda likha hai; unke beech dashed vermilion curves (nodal lines, jahan crest trough se milta hai) par lambda over 2 likha hai. Step 3: right mein ek vertical screen hai jiske paas intensity ka graph hai, jo wahan bahar ubharta hai jahan green line screen se milti hai aur wahan screen ko chhoota hai jahan dashed line milti hai.

DerivationKisi bhi phase difference ke liye intensity

In extremes ke beech intensity smoothly badalti hai. Maan lo har source akele $P$ par amplitude $a$ deta hai; sources unke separation ke comparison mein bahut door hain, isliye dono amplitudes barabar hain. $P$ par displacements hain

y_{1} = a \cos ω t

y_{2} = a \cos (ω t + ϕ)

Identity $\cos α + \cos β = 2 \cos \frac{α - β}{2} \cos \frac{α + β}{2}$ se inhe jodne par milta hai

y = A \cos ⁣ (ω t + \frac{ϕ}{2})

A = 2 a \cos \frac{ϕ}{2}

Yeh phir se angular frequency $ω$ wala oscillation hai, ab amplitude $A = 2 a \cos (ϕ / 2)$ ke saath. Wave ki intensity uske amplitude ke square ke hisaab se chalti hai, isliye agar akela ek source $I_{0} \propto a^{2}$ deta hai,

I = 4 I_{0} \cos^{2} \frac{ϕ}{2}

(10.11)

$ϕ = 0, \pm 2 π, \pm 4 π, \dots$ (path difference $n λ$ ) ke liye yeh $4 I_{0}$ deta hai, ek source ka chaar guna, do guna nahi. $ϕ = \pm π, \pm 3 π, \dots$ (path difference $(n + \frac{1}{2}) λ$ ) ke liye yeh zero deta hai (Fig. 10.9).

1/3Step mein: amplitude $2 a$ , intensity $4 I_{0}$ .

Fig. 10.9Amplitudes judte hain; intensity amplitude ke square ke hisaab se chalti hai. Step mein:

4 I_{0}

. Quarter cycle ke farq par:

2 I_{0}

. Half cycle ke farq par: kuch nahi.NCERT Eq. (10.11) · p. 264

Upar: same amplitude ki ek solid blue wave y1 aur ek dashed orange wave y2. Neeche: unka sum, ek moti black wave. Right: ek yellow bar intensity dikhata hai. Step 1, phase difference zero: dono waves ek doosre ke upar hain, sum ka amplitude do guna hai, aur bar poora bhara hai, 4 I nought. Step 2, phase difference pi over 2: orange wave ek quarter wavelength khisak jaati hai, sum ka amplitude 1.41 guna hai, aur bar aadha reh jaata hai, 2 I nought. Step 3, phase difference pi: orange wave half wavelength khiski hai, crest trough ke saamne, sum ek flat line hai aur bar khaali hai.

Ab maan lo dono prongs alag motors se chalte hain jo kabhi step mein aate hain, kabhi bahar. Tab kisi bhi point par phase difference $ϕ$ saari values mein ghoomta rehta hai, aur Fig. 10.10 ka pattern tank mein aage-peeche sarakta hai. Agar $ϕ$ itni tezi se badle ki hum dekh na paayein, to humein sirf $\cos^{2} (ϕ / 2)$ ka time average dikhta hai, jo $\frac{1}{2}$ hai:

\overset{ˉ}{I} = 4 I_{0} \times \frac{1}{2} = 2 I_{0}

(10.12)

Aise sources incoherent hote hain: unki intensities bas jud jaati hain, aur koi fringes nahi dikhti. Garage ke darwaze par padti car ki do headlights aisa hi karti hain. Jahan dono beams padti hain, wahan darwaza ek beam ke comparison mein do guna bright hota hai, ek jaisa, bina kisi pattern ke nishaan ke.

3/3Tez random jumps ka average ek uniform $2 I_{0}$ ban jaata hai.

Fig. 10.10Agar phase difference fixed rahe to fringes ruki rehti hain. Agar yeh uchhalta rahe, to fringes aage-peeche sarakti hain, aur aankh sirf unka average dekhti hai, har jagah

2 I_{0}

.NCERT Eq. (10.12) · p. 264

Screen par position ke against intensity ka graph, jismein 4 I nought aur 2 I nought par dashed reference lines hain. Step 1: ek blue curve 0 aur 4 I nought ke beech upar-neeche hoti hai, label kehta hai steady phase difference. Step 2: pehle wala curve halka aur dashed dikhta hai, aur usi shape ka ek orange curve side mein khiska hai, label kehta hai ki phase jumps se fringes sarakti hain. Step 3: kai halke khiske curves ek doosre ke upar hain aur 2 I nought par ek moti green horizontal line unka average dikhati hai, label kehta hai ki aankh average dekhti hai.

Example

Do loudspeakers, ek tone

Paas-paas rakhe do loudspeakers ko ek hi amplifier $680 H z$ tone ke saath in phase chalata hai, aur sound $340 m / s$ se chalta hai. Maan lo har speaker akela har seat par same intensity $I_{0}$ deta hai. (Asal mein paas wala speaker thoda zyada loud hota hai, isliye quiet spot bilkul silent nahi hota.) Us seat par intensity nikaalo jo (a) ek speaker se $3.00 m$ aur doosre se $3.75 m$ door hai, (b) $3.00 m$ aur $4.00 m$ door hai, (c) $3.000 m$ aur $3.125 m$ door hai.

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

Quick check

4 mein se 1

Do coherent sources mein se har ek us point par intensity

I_{0}

deta hai jahan path difference

λ / 6

hai. Resultant intensity kitni hai,

I_{0}

ke multiple mein?

× I₀

Answer dekho

Answer 3 × I₀

ϕ = 2 π (λ / 6) / λ = π / 3

, isliye

I = 4 I_{0} \cos^{2} (π / 6) = 4 I_{0} \times \frac{3}{4} = 3 I_{0}

10.5 Light waves ka interference aur Young's experiment

Ripple-tank wala experiment light ke saath try karo: do pinholes mein se har ek ke peeche ek chhota green LED rakho aur unke aage ek screen dekho. Woh ek jaisi roshan hai, koi fringes nahi. Wajah yeh hai ki aise sources light kaise banate hain. Light anginat atoms se aaye chhote wave trains ka ek mishran hai, aur uska phase randomly bhatakta rehta hai; green LED ke liye yeh har kuch times $10^{- 14} s$ mein lagbhag poori tarah gadbad ho jaata hai. Ek picture register karne ke liye tumhari aankh ko jo second ka beesvaan hissa chahiye, usmein koi bhi fringe pattern ek million million se zyada baar badal jaata. Do independent sources hamesha incoherent hote hain.

Young ne iska hal behtar lamps dhoondhkar nahi, balki ek source ko do mein baantkar nikala. Fig. 10.11 mein ek patli slit $S$ lamp ke ek chhote hisse se light guzarne deti hai, aur woh light millimetre ke ek hisse jitni door rakhi do slits $S_{1}$ aur $S_{2}$ tak pahunchti hai. Ab dono slits ek hi amplifier se jude do loudspeakers jaisi hain: lamp chahe jitna bhi bekaboo behave kare, dono ko same wave milti hai, isliye unke phases ka difference kabhi nahi badalta. Aaj laser, jiski light lambi distances tak step mein rehti hai, is experiment ko aasaan bana deta hai. Slits ke aage screen par bright aur dark bands dikhte hain jinhe interference fringes (व्यतिकरण) kehte hain.

4/4Jahan extra path $n λ$ hai wahan bright; inke beech ki doori $β = λ D / d$ hai.

Fig. 10.11Ek slit do ko light deti hai, isliye dono slits coherent hain.

P

tak extra path

d \sin θ \approx x d / D

hai, aur screen par barabar doori wali fringes dikhti hain.NCERT Fig. 10.12 · p. 265

Left se right: ek yellow lamp, ek slit S wala barrier, distance d par do slits S1 aur S2 wala barrier, aur distance D par ek screen. Step 1: blue arcs S se dono slits tak aur har slit se screen ki taraf phailte hain. Step 2: S1 se ek blue line aur S2 se ek orange line screen par point P par milti hain, jo central point O se height x upar hai, aur slits se O tak ek dashed line hai. Step 3: ek bada kiya gaya inset dono slits ko dikhata hai jinse angle theta par parallel rays nikalti hain; S1 se ek dashed line neeche wali ray se milti hai aur ek chhota mota vermilion segment kaat-ti hai, path difference d sine theta, lagbhag x d over D. Step 4: screen ke paas ek strip barabar doori par yellow bright fringes aur unke beech dark gaps dikhati hai, aur paas-paas ki bright fringes ke beech ki doori beta marked hai.

DerivationFringes kahan padti hain

Maan lo slits ke beech distance $d$ hai aur screen distance $D$ par hai, jahan $D ≫ d$ . Screen par point $O$ dono slits se barabar door hai. $O$ se height $x$ upar ek point $P$ lo. Pythagoras se, $(S_{2} P)^{2} = D^{2} + (x + d / 2)^{2}$ aur $(S_{1} P)^{2} = D^{2} + (x - d / 2)^{2}$ . Ghataane par $D^{2}$ terms cancel ho jaate hain aur $(S_{2} P)^{2} - (S_{1} P)^{2} = 2 x d$ .

Left side ke factors $(S_{2} P - S_{1} P) (S_{2} P + S_{1} P)$ hain, aur jab $x$ aur $d$ , $D$ se bahut chhote hon, to $S_{2} P + S_{1} P \approx 2 D$ . Isliye path difference $Δ = S_{2} P - S_{1} P$ hai

Δ \approx \frac{x d}{D}

Isi ko $Δ = d \sin θ$ bhi likh sakte hain, jahan $θ$ woh angle hai jis par slits se $P$ dikhta hai, kyunki chhote angles ke liye $\sin θ \approx \tan θ = x / D$ . Bright fringes ke liye $Δ = n λ$ aur dark fringes ke liye $Δ = (n + \frac{1}{2}) λ$ chahiye, isliye $n = 0, \pm 1, \pm 2, \dots$ ke liye

Bright fringes

x_{n} = n \frac{λ D}{d}

(10.13)

Dark fringes

x_{n} = (n + \frac{1}{2}) \frac{λ D}{d}

(10.14)

Paas-paas ki bright fringes, aur paas-paas ki dark fringes bhi, same distance se alag hoti hain, jise fringe width kehte hain:

Fringe width

β = x_{n + 1} - x_{n} = \frac{λ D}{d}

(10.13) aur (10.14) se chaar baatein nikalti hain:

Fringes barabar doori par hoti hain, bright aur dark baari-baari se, aur centre $O$ par ek bright fringe hoti hai, jahan path difference zero hai.
$β \propto λ$ : red fringes blue wali se chaudi hoti hain. White light mein central fringe white hoti hai, kyunki wahan har colour ka path difference zero hai, aur uske dono taraf kuch coloured fringes hoti hain jo jaldi hi aapas mein ghul-mil jaati hain.
$β \propto 1 / d$ : slits ko paas laane se fringes phail jaati hain. Isiliye slits bahut paas honi chahiye, aam taur par millimetre ke ek hisse jitni door.
Poore apparatus ko refractive index $n$ wale liquid mein dubone se wavelength ghatkar $λ / n$ ho jaati hai, isliye fringes bhi usi factor se paas aa jaati hain.

InteractiveYoung's double-slit lab

Ek simulated screen chune gaye colour mein do slits ki fringes dikhati hai, aur uske neeche minus 15 se plus 15 millimetres tak position ke against intensity ka graph hai. Sliders wavelength ko 400 se 700 nanometres, slit separation ko 0.10 se 1.00 millimetre aur screen distance ko 0.5 se 3.0 metres tak set karte hain, aur ek menu har slit ki width set karta hai (ideal, 0.02, 0.05 ya 0.10 millimetre), jo ek single-slit envelope jodta hai. Ek switch coherent slits ki jagah do alag lamps laga deta hai, jisse ek jaisa 2 I nought milta hai, aur ek button khisakte pattern ke snapshots ek ke upar ek dikhata hai. Ek probe slider screen par ek point chunta hai aur uska path difference nanometres mein aur wavelengths mein, phase difference aur intensity batata hai. Readouts fringe width aur angular fringe width dete hain.

Fig. 10.12Colour, slit separation aur screen distance badlo, aur fringe width dekho. Phir slits ko ek width do, ya do alag lamps par switch karo.

History seNCERT Thomas Young · p. 266

Thomas Young (1773–1829)

Kaha jaata hai ki Young do saal ki umar mein dhadalle se padh lete the, aur teenager hote-hote unhone ek darjan languages seekh li thin. Unhone physician ki training li aur London mein practice ki, aur unki pehli discoveries aankh ke baare mein thin: uska lens focus karne ke liye shape kaise badalta hai, aur yeh idea ki colour vision ke liye sirf teen tarah ke receptors kaafi hain. Mechanics mein unka naam Young's modulus mein zinda hai, jo kisi material ki stiffness ka measure hai. 1801–1803 ke unke interference experiments ne wave theory ko phir se zinda kiya, halaanki Britain mein critics ne pehle unka mazaak udaaya. Unhone Rosetta Stone ke Egyptian hieroglyphs padhne mein bhi pehli asli tarakki ki, Champollion ke unhe poori tarah decipher karne se pehle.

Example

Green laser se fringes

Green laser ( $λ = 532 n m$ ) ki light $0.40 m m$ door rakhi do slits par padti hai, aur fringes $1.6 m$ door ek deewar par dikhti hain. Nikaalo (a) fringe width, (b) central bright fringe se teesri dark fringe tak ki distance, (c) angular fringe width, aur (d) fringe width agar poora apparatus water ( $n = 1.33$ ) mein duba diya jaaye.

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

10.6 Diffraction

Yaad hai tube light ke saamne lagbhag milakar pakdi do ungliyon ke beech tumhe jo dark lines dikhi thin (Section 10.1)? Patle gap se nikalti light seedhi aage nahi jaati, jaisa ray optics kehta hai. Woh phail jaati hai, aur phaili hui light bright aur dark bands banati hai. Kaafi chhote source ke saath tum yahi cheez kisi bhi shadow ke edge par dekh sakte ho, jo kabhi bilkul sharp nahi hota. Obstacles ke around aur openings se hokar waves ke is phailne ko diffraction (विवर्तन) kehte hain.

Har tarah ki wave diffract hoti hai. Harbour wall ke end se guzarti samundar ki lehrein mudkar uske peeche ke shaant paani mein aa jaati hain, aur agli gali mein bajta band building ke kone ke paar tum tak pahunch jaata hai. Light bhi utni hi aasaani se diffract hoti hai, lekin uski wavelength sound se lagbhag das lakh guna chhoti hai, isliye effects sirf millimetre ke ek hisse jitni openings aur obstacles ke liye dikhte hain. Isiliye roz ke shadows sharp dikhte hain, aur isiliye ray optics itna achha kaam karta hai.

Diffraction optical instruments par ek pakki limit bhi lagata hai. Finite size ka lens ya mirror light ke ek point ko bilkul point par focus nahi kar sakta: woh ek chhota sa diffraction blur banata hai, aur us blur ke size se zyada paas rakhe do objects alag nahi pehchaane ja sakte, lens chahe kitna bhi achha bana ho. Isliye microscope ya telescope jo sabse baareek detail dikha sakta hai, woh light ki wavelength se tay hoti hai. Openings ke regular pattern se bhi diffraction dikhta hai: chhatri ke kapde se door ki street light ko dekho, aur dhaagon ke beech ka har gap light ko rangeen kinaron wale spots ke ek grid mein phaila deta hai.

10.6.1 Single slit

Monochromatic light ki ek parallel beam ko width $a$ wali patli slit par seedha padne do, aur door ki screen par (ya slit ke baad rakhe lens ke focus par) light dekho. Jo dikhta hai woh slit jitni chaudi ek sharp patti nahi hai. Screen ke beech mein ek chaudi bright band hoti hai, aur har taraf kaafi dheemi bands ki ek row hoti hai, jo dark gaps se alag hoti hain aur doori ke saath halki padti jaati hain. Slit jitni patli, poora pattern utna zyada phailta hai.

Slit ko kai chhote wavelet sources ki ek row maano (phir se Huygens). Aane wala wavefront slit ke aar-paar flat pada hai, isliye har source same phase par shuru hota hai; farq bas itna hai ki har ek us point se kitni door hai jahan hum dekh rahe hain. Direction $θ$ mein, neeche wale edge ki wavelet upar wale edge ki wavelet se $a \sin θ \approx a θ$ zyada chalti hai, aur beech ki wavelets is range mein barabar phaili hain (Fig. 10.13).

DerivationDark fringes kahan padti hain

Seedhe aage ( $θ = 0$ ): har wavelet same distance chalti hai, isliye sab in phase pahunchti hain aur jud jaati hain. Yahi bright central maximum hai.

$a θ = λ$ par: slit ko upar wale half aur neeche wale half mein baanto. Upar wale half ke har point ka ek partner theek $a / 2$ neeche hai, aur dono partners ke paths mein $λ / 2$ ka farq hai. Har pair cancel hota hai, isliye intensity zero hai: pehli dark fringe.

$a θ = 2 λ$ par: slit ko quarters mein baanto. Pehla aur doosra quarter usi tarah cancel hote hain, aur teesra aur chautha bhi, isliye phir dark hai. General mein minima yahan hote hain

Single-slit minima

θ \approx \frac{n λ}{a}

jahan

$n$: $\pm 1, \pm 2, \pm 3, \dots$ (zero nahi)

$a θ = \frac{3}{2} λ$ par: slit ko thirds mein baanto. Do thirds cancel ho jaate hain, aur bache hue third ki light bach jaati hai, isliye pehle do minima ke lagbhag beech mein ek secondary maximum hota hai. Yeh weak hota hai, kyunki slit ka sirf ek third contribute karta hai aur uski apni wavelets bhi thodi out of step hoti hain. Poori calculation pehle secondary maximum ke liye central intensity ka lagbhag $4.7 %$ aur agle ke liye $1.6 %$ deti hai. Secondary maxima $θ \approx \pm (n + \frac{1}{2}) λ / a$ ke paas hote hain, $n = 1, 2, \dots$ ke saath

2/4Dono halves pair-by-pair cancel hote hain: pehla minimum.

Fig. 10.13Wavelets ke pairs banao: poore halves, quarters, sixths… cancel hokar dark fringes dete hain. Angles bada-chadha kar dikhaye gaye hain.NCERT Fig. 10.14 · p. 267

Left: plane blue wavefronts ek barrier ki slit par aate hain; slit ke aar-paar aath dots wavelet sources mark karte hain, aur unse grey parallel rays ek direction mein nikalti hain. Slit ke paas ek coloured bar dikhata hai ki slit kaise baanti gayi hai. Right: angle ke against intensity ka graph, ek lambe central peak aur chhote side peaks ke saath, aur ek red dot abhi ki direction mark karta hai. Step 1: rays seedhe aage jaati hain aur bar ek green block hai, sab step mein: bright; dot central peak ke upar hai. Step 2: rays itni jhukti hain ki edges ke beech path difference, ek mota red segment, ek wavelength hai; bar ek blue upar wale half aur ek orange neeche wale half mein bata hai jo cancel hote hain: dark; dot pehle zero par hai. Step 3: jhukaav dedh wavelength ka path difference deta hai; bar thirds mein hai, do cancel hote hain aur ek green third bachta hai: halka; dot pehle chhote side peak par hai. Step 4: path difference do wavelengths hai; bar chaar baari-baari wale quarters mein hai jo pairs mein cancel hote hain: dark; dot doosre zero par hai.

Central maximum $- λ / a$ se $+ λ / a$ tak jaata hai, isliye uski angular width $2 λ / a$ hai, baaki bands ki do guna. Distance $D$ par rakhi screen par yeh $2 λ D / a$ chauda hota hai. Slit ko patla karne se pattern phailta hai; use wavelength se bahut chauda karne par pattern simatkar slit ki ek sharp image ban jaata hai, aur ray optics kaam sambhaal leta hai.

Example

Central band ki width

$600 n m$ wavelength ki light $0.10 m m$ chaudi slit par normally padti hai, aur pattern $2.0 m$ door screen par dikhta hai. Nikaalo (a) pehle minimum ka angle, (b) screen par central maximum ki width, aur (c) yahi width agar iski jagah $450 n m$ wavelength ki blue light use ki jaaye.

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

Compare

Interference aur diffraction

Interference (double slit mein)

Kuch coherent sources (do slits) ki light ko joda jaata hai
Saari fringes same width ki, $λ D / d$
Bahut patli slits ke liye lagbhag barabar brightness wali bright fringes
Spacing slit separation $d$ se tay hoti hai

Diffraction (single slit mein)

Ek opening ke aar-paar wavelets ke continuous band ki light ko joda jaata hai
Central band baaki sab se do guna chauda
Centre se door brightness tezi se ghatti hai
Phailaav opening width $a$ se tay hota hai

Dono effects mein superposition hi kaam kar raha hai, aur unke beech koi gehri physical line nahi hai. Richard Feynman ne kaha tha ki farq sirf bolne ke tareeke ka hai: jab kuch sources jodte hain to hum interference kehte hain, aur jab bahut saare jodte hain to diffraction.

Asli double-slit experiment mein har slit ki apni ek width hoti hai, isliye dono effects ek saath kaam karte hain. Har slit apni light ko single-slit pattern mein phailati hai, aur us phailaav ke andar dono slits interfere karti hain. Result hai Young's fringes, $λ D / d$ ki spacing ke saath, jinki brightness single-slit pattern ko envelope ki tarah follow karti hai (Fig. 10.14). Jahan interference maximum diffraction minimum par padta hai, wahan woh fringe gaayab hoti hai.

Fig. 10.14Asli double slit: single-slit envelope ke andar interference fringes. Aisi slits ke liye banaya gaya jinke centres chaar slit-widths door hain, isliye 4th-order fringes envelope ke pehle zeros par padti hain aur gaayab hain.NCERT 10.6.1 · p. 268

Angle ke against intensity ke teen graphs, ek ke upar ek, same horizontal scale ke saath. Upar, ek slit: ek chauda central peak aur chhote side peaks. Beech mein, do ideal slits: barabar unchi, barabar doori wali patli peaks ki ek row. Neeche, do asli slits: wahi patli peaks, lekin unki unchaai upar wale graph ki shape ko follow karti hai, jo ek dashed envelope ki tarah bani hai; har taraf ki chauthi peak, jo envelope ke zero par padti hai, gaayab hai aur us par aisa likha hai.

Example

Central envelope ke andar fringes

Do slits, har ek $0.060 m m$ chaudi, ke centres $0.30 m m$ door hain. Central diffraction maximum ke andar kitni bright interference fringes hain?

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

10.6.2 Single slit diffraction pattern dekhna

Tum ghar ya school laboratory mein milne wali cheezon se single-slit pattern dekh sakte ho. Tumhe bas ek patli, adjustable slit aur light ki ek patli, bright line chahiye.

Ghar par try karoNCERT 10.6.2 · p. 268

Calliper gap se diffraction

Tumhe chahiye

ek vernier calliper, ya do nayi razor blades
seedhe filament wala ek clear bulb, ya door ki tube light
red aur blue cellophane sheets (optional)

Pehle guess karo

Jaws ke beech ka gap band karte hue tumhe dikhne wale bands

aur door-door phail jaate hainaur paas-paas simat jaate hainwaise hi rehte hain, bas dheeme ho jaate hain

Vernier calliper ke jaws ko itna band karo ki woh lagbhag chhoo jaayein, lagbhag $0.1 m m$ ka gap chhodkar (ya do nayi razor blades ke edges aamne-saamne pakdo). Check karo ki gap apni poori length mein ek jaisa hai.
Seedhe filament wale clear bulb, ya door ki tube light se kuch metres door khade ho. Gap ko ek aankh ke bilkul saamne, filament ke parallel pakdo, aur usmein se light ko dekho. Agar tum aam taur par chashma pehente ho to pehen lo.
Gap ko dheere-dheere kholo aur band karo. Phir red cellophane se dekho, aur phir blue se.

Table 10.2Tumhe kya dikhta hai

	Tumhe kya dikhta hai
Gap lagbhag 0.1 mm
Gap aur patla
Gap aur chauda
Red cellophane se
Blue cellophane se

Tumhari entries apne aap save hoti hain.

Tumhe ek bright central band dikhta hai, jiske dono taraf halke bands hain, jo white light mein apne kinaron par rangeen hote hain, kyunki central ke alawa har band ki position wavelength par depend karti hai. Gap patla karne se pattern phail jaata hai. Red cellophane se bands blue ke comparison mein chaude dikhte hain, kyunki $λ_{red} > λ_{blue}$ . Tumhari aankh ka lens single-slit set-up wale focusing lens ka kaam karta hai, isliye tum jo dekhte ho woh angle ke hisaab se phaila pattern hai, jo tumhari retina par image banta hai.

Ab usi lamp ke saamne do ungliyan lagbhag milakar pakdo, jaise chapter ki shuruaat mein kiya tha. Gap mein dikhne wali dark lines yahi diffraction pattern hain, ek aisi slit se bani jiski width tum ungliyan dabaakar badal sakte ho.

Yeh try karo

Laser se baal naapo

Tumhe chahiye

ek low-power red laser pointer
baal ka ek tukda
card ka ek tukda aur tape
ek metre rule aur kam se kam 2 m door ek deewar

Pehle guess karo

Beam mein baal rakhne par tumhe deewar par kya dikhega?

Ek sharp shadow lineBright spots ki ek row, slit pattern jaisiKuch nahi: baal bahut zyada patla hai

Card ke tukde mein ek ched ke aar-paar ek baal tape se chipkao taaki woh seedha aur vertical rahe.
Ek low-power red laser pointer ( $λ \approx 650 n m$ ) ko baal se hokar kam se kam $2 m$ door deewar par chamkao. Baal se deewar tak ki distance $D$ naapo.
Pattern ke bright centre ke dono taraf ke do dark gaps ke beech ki distance $w$ naapo. Phir $a = 2 λ D / w$ se baal ki width calculate karo.

Table 10.3Tumhara measurement

	Value
Deewar tak distance $D$ (m)
Central band ki width $w$ (m)
Baal ki width $a = 2 λ D / w$ (µm)

Tumhari entries apne aap save hoti hain.

Yeh ek check karne laayak baat hai. Do slits ke liye intensity $0$ aur $4 I_{0}$ ke beech upar-neeche hoti hai, aur ek fringe par uska average $2 I_{0}$ hai: bilkul utna jitna dono slits alag-alag detin. Isliye dark fringe mein light nasht nahi hoti, balki kahin aur bhej di jaati hai, paas ki bright fringes mein. Single slit ke liye bhi yahi hisaab-kitaab lagta hai, jahan poore pattern mein phaili energy us energy ke barabar hai jo opening se guzri, jaisa energy conservation maangta hai.

Quick check

4 mein se 1

Ek single slit ko patla kiya jaata hai. Screen par central bright band

chauda ho jaata haipatla ho jaata haiwidth wahi rehti hai lekin dheema ho jaata haido bands mein bat jaata hai

Answer dekho

Answer (A) chauda ho jaata hai

Iski angular half-width

λ / a

hai, isliye chhota

a

chauda band deta hai. (Yeh dheema bhi hota hai, kyunki kam light guzarti hai, lekin iski width badhti hai.)

10.7 Polarisation

Ek lambi rassi ka ek end khambe se baandho aur free end ko regularly upar-neeche hilao. Rassi ke along ek wave chalti hai (ise $x$ direction kaho), aur rassi ka har tukda sirf upar-neeche, $y$ ke along, hilta hai:

y (x, t) = a \sin (k x - ω t)

(10.15)

Yahan $a$ amplitude hai, $ω = 2 π ν$ angular frequency hai, aur

k = \frac{2 π}{λ}

(10.16)

angular wave number hai. Rassi ke kisi bhi ek tukde ko dekho: woh ek fixed line ke along upar-neeche hilta hai, hamesha wave ke chalne ki direction ke right angle par, isliye yeh transverse wave hai. Jis transverse wave mein har particle ek fixed line par hi rehta hai, use linearly polarised (रैखिकतः ध्रुवित) kehte hain. Yahan line $y$ axis hai, isliye yeh $y$ -polarised wave hai; kyunki poori rassi $x$ – $y$ plane mein rehti hai, aisi wave ko plane polarised bhi kehte hain. Iski jagah rassi ko side to side hilao to tumhe $z$ -polarised wave milti hai:

z (x, t) = a \sin (k x - ω t)

(10.17)

Agar tum hilaane ki direction second mein kai baar randomly badalte raho, to vibration ab bhi rassi ke right angle par hai lekin uski koi fixed direction nahi: wave unpolarised hai. Ab rassi ko ek board ke vertical slot se guzaaro (Fig. 10.15). Upar-neeche wali wave guzar jaati hai, side-to-side wali ruk jaati hai, aur unpolarised wave vertically polarised hokar nikalti hai, jismein sirf uska upar-neeche wala part bachta hai.

4/4Polaroid $𝐄$ ka sirf woh component guzarne deta hai jo uske pass-axis ke along ho.

Fig. 10.15Slot transverse vibration ka sirf woh part guzarne deta hai jo uske along ho. Polaroid light ke electric field ke saath yahi karta hai.NCERT Fig. 10.17 · p. 269

Teen columns, har ek chalne ki direction mein dekha gaya end-on view, arrows se jude hue: kya aata hai, filter, aur kya nikalta hai. Step 1: aane wali rassi upar-neeche vibrate karti hai (ek purple vertical double arrow); filter ek lakdi ka board hai jismein vertical slot hai; nikalne wali wave ab bhi upar-neeche vibrate karti hai, label kehta hai ki guzar gayi. Step 2: aane wali rassi side to side vibrate karti hai (ek horizontal double arrow); kuch nahi nikalta, label kehta hai ki ruk gayi. Step 3: aane wali vibration random hai, chhe directions mein double arrows ki tarah dikhayi gayi; sirf ek chhota vertical double arrow nikalta hai, label kehta hai sirf upar-neeche. Step 4: rassi ki jagah light hai; chhe directions mein orange double arrows unpolarised light ka electric field dikhate hain, filter ek polaroid disc hai jismein horizontal chain lines aur ek dashed vertical pass-axis hai, aur ek akela vertical orange arrow nikalta hai, label kehta hai polarised.

Light bhi isi tarah behave karti hai, aur yahi humara saboot hai ki light waves transverse hain. Light wave mein electric field $𝐄$ hamesha chalne ki direction ke perpendicular hota hai, aur jis direction mein $𝐄$ oscillate karta hai use polarisation (ध्रुवण) ki direction kehte hain. Sun, bulb ya candle ki light unpolarised hoti hai: uske atoms independently emit karte hain, isliye $𝐄$ ki direction tezi se aur randomly badalti rehti hai, halaanki hamesha ray ke perpendicular plane ke andar hi.

Polaroid light par waisa hi kaam karta hai jaisa slot rassi par. Har polaroid ke plane mein ek khaas direction hoti hai, uska pass-axis: $𝐄$ ka is direction ke along wala component guzar jaata hai, aur iske right angle wala component absorb ho jaata hai. Is direction ko khaas material banata hai. Ek typical sheet polyvinyl alcohol ki hoti hai, jise is tarah stretch kiya jaata hai ki uske long-chain molecules line mein aa jaayein, aur iodine se rang diya jaata hai taaki electrons chains ke along chal sakein. Chains ke along electric field unmein chhote currents chalata hai aur apni energy kho deta hai; chains ke aar-paar wala field aisa nahi kar sakta, aur guzar jaata hai. Isliye pass-axis chains ke perpendicular hota hai. Edwin Land ne 1929 mein polarising sheets invent kin, aur yeh stretched, iodine wali kism 1938 se use ho rahi hai.

Ek lamp ke saamne ek polaroid rakho. Randomly oriented $𝐄$ mein se sirf pass-axis ke along wala component guzarta hai. Saari directions ka average lene par yeh aadhi intensity hai, sheet ko kisi bhi taraf ghumao, isliye use ghumaane se koi dikhne wala farq nahi padta. Lekin nikalne wali light ab pass-axis ke along linearly polarised hai.

Ek doosra polaroid jodo. Agar uska pass-axis pehle wale ke parallel hai, to polarised light seedhi guzar jaati hai. Use ghumao, aur light dheemi hoti hai; $90^{\circ}$ par (crossed polaroids) lagbhag kuch nahi guzarta. Fig. 10.16 iski wajah dikhata hai. Maan lo pehli sheet se nikalne wali light ka amplitude $E_{0}$ aur intensity $I_{0}$ hai, aur doosra pass-axis pehle ke saath angle $θ$ banata hai. Sirf doosre axis ke along wala component $E_{0} \cos θ$ transmit hota hai, aur intensity amplitude ke square ke hisaab se chalti hai.

LawMalus' law

Jab intensity $I_{0}$ wali linearly polarised light aise polaroid par padti hai jiska pass-axis polarisation ki direction ke saath angle $θ$ banata hai, to transmitted intensity $I_{0} \cos^{2} θ$ hoti hai.

Malus' law

I = I_{0} \cos^{2} θ

(10.18)

2/3 $θ = 60^{\circ}$ : guzarne wala field $E_{0} \cos 60^{\circ} = E_{0} / 2$ hai, isliye $I = I_{0} / 4$ .

Fig. 10.16Sirf component

E_{0} \cos θ

guzarta hai, isliye intensity

I_{0} \cos^{2} θ

hai:

60^{\circ}

par ek chauthai, crossed hone par kuch nahi.NCERT Fig. 10.18 · p. 271

Left: pehla polaroid, ek purple disc jismein vertical dashed pass-axis hai aur usse nikalne wali light ke electric field E nought ke liye ek orange vertical double arrow hai. Beech mein: doosra polaroid apne dashed pass-axis ke saath, aane wala field halka dikhaya gaya hai, aur jo part yeh guzarne deta hai uske liye ek green double arrow. Right: ek yellow intensity bar. Step 1: doosra axis vertical hai, poora field guzarta hai aur bar bhara hai, I nought. Step 2: doosra axis 60 degrees ghumaya gaya hai; ek dashed line field ko us par project karti hai, green arrow aadha lamba hai, E nought cos 60 degrees, aur bar ek chauthai reh jaata hai, I nought over 4. Step 3: axis horizontal hai, pehle ke saath crossed; koi green arrow nahi bachta aur bar khaali hai.

Isliye do sheets ek dimmer bana deti hain: unmein se ek ko ghumaane se transmitted intensity lamp ki aadhi intensity (axes parallel) aur zero (axes crossed) ke beech kahin bhi set ho jaati hai. Jab doosri sheet poora chakkar ghoomti hai, to light do maxima se guzarti hai, $θ = 0^{\circ}$ aur $180^{\circ}$ par, aur do minima se, $90^{\circ}$ aur $270^{\circ}$ par.

InteractivePolaroid lab

Ek lamp unpolarised light bhejta hai, jo arrows ke ek star ki tarah bani hai, aur yeh light teen tak polaroid discs se hokar, jinke pass-axis ek line ki tarah bane hain, ek light meter tak jaati hai. Sheets ke beech ek arrow electric field ki direction aur size dikhata hai. Sliders pehli sheet, optional beech wali sheet aur aakhri sheet ko ghumaate hain, aur check boxes beech wali aur aakhri sheets ko jodte ya hataate hain. Readouts har sheet ke baad intensity ko lamp ki intensity ke percentage mein dete hain, Malus' law ki calculation likhi hui. Ek graph aakhri sheet ke 0 se 360 degrees tak ke angle ke against transmitted intensity plot karta hai, abhi ki setting ke liye ek dot ke saath. Play button aakhri sheet ko dheere-dheere ghumata hai, aur ek preset button bahar wali sheets ko crossed kar deta hai.

Fig. 10.17Har sheet ko ghumao aur uske baad ki intensity padho. Ek akeli sheet try karo, do crossed sheets, aur phir crossed pair ke beech ek teesri sheet khiska do.

Example

Do polaroids

$50 W / m^{2}$ intensity wali unpolarised light do polaroids par padti hai jinke pass-axes ek doosre se $35^{\circ}$ par hain. (a) Kitni intensity nikalti hai? (b) $10 W / m^{2}$ nikalne dene ke liye doosri sheet ko pehli ke relative kis angle tak ghumaana hoga?

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

Example 10.2NCERT Example 10.2 · p. 271

Crossed polaroids ke beech ghoomti sheet

Do polaroids $P_{1}$ aur $P_{3}$ crossed hain, isliye akele woh koi light nahi guzarne dete. Unke beech ek teesra polaroid $P_{2}$ steady 5 revolutions per second se ghumaya jaata hai; $t = 0$ par uska pass-axis $P_{1}$ ke pass-axis ke parallel hai. $80 W / m^{2}$ intensity wali unpolarised light $P_{1}$ par padti hai.

(a) Jab $P_{2}$ ka axis $P_{1}$ ke axis ke saath angle $θ$ par ho, to $P_{3}$ se nikalne wali intensity nikaalo. (b) $t = 1 / 60 s$ par yeh kitni hai? (c) Nikalne wali light har second kitni baar sabse bright hoti hai, aur tab kitni bright hoti hai? (d) Ek slow light meter kitni average intensity padhta hai?

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

Ghar par try karo

Sunglasses aur ek screen

Tumhe chahiye

polarising sunglasses (aksar un par 'polarised' likha hota hai)
ek phone, laptop ya calculator ka display

Pehle guess karo

Sunglasses ke ek lens se screen ko dekhte hue, tum glasses ko dheere-dheere poora chakkar ghumaate ho. Tumhe kya ummeed hai?

Screen barabar bright rehti haiEk chakkar mein screen do baar dark hoti haiEk chakkar mein screen ek baar dark hoti hai

Phone, laptop ya calculator ka display on karo aur sunglasses ko uske saamne pakdo, ek lens se dekhte hue.
Glasses ko dheere-dheere poora chakkar ghumao, aur note karo ki kin positions par screen sabse dark dikhti hai.
Ab glasses se kisi chamakdaar table ya paani ke gaddhe se reflect hui light dekho, aur apna sir ek taraf jhukao.

Quick check

4 mein se 1

100 W / m^{2}

intensity wali unpolarised light do polaroids se guzarti hai jinke pass-axes ek doosre se

60^{\circ}

par hain. Kitni intensity nikalti hai,

W / m^{2}

mein?

W/m²

Answer dekho

Answer 12.5 W/m²

Pehli sheet

50 W / m^{2}

guzarne deti hai; doosri

50 \cos^{2} 60^{\circ} = 50 \times 0.25 = 12.5 W / m^{2}

guzarne deti hai.

Recap

Summary aur key points

Light ke particle aur wave models water mein ulti speeds predict karte hain; naapne par light wahan dheemi nikalti hai, jaisa wave model kehta hai (Section 10.1).
Wavefront constant phase ki surface hai, aur rays iske perpendicular hoti hain. Point source spherical wavefronts deta hai; door ka source lagbhag plane wavefronts deta hai (Section 10.2).
Huygens' principle: wavefront ka har point secondary wavelets ka source hai, aur unka forward envelope time $τ$ baad ka wavefront hai (Principle).
Huygens' construction se Snell's law milta hai, $n_{1} \sin i = n_{2} \sin r$ , jahan $n = c / v$ , aur yeh predict karta hai ki jis medium mein light normal ki taraf mudti hai usmein woh dheemi hai (Section 10.3.1).
Boundary par frequency nahi badalti, jabki speed aur wavelength same factor se badalte hain ((10.7)).
Rarer medium mein jaati light normal se door mudti hai. Critical angle ke aage, $\sin i_{c} = n_{2} / n_{1}$ , koi refracted wave nahi ban sakti aur light poori tarah reflect ho jaati hai (Section 10.3.2).
Reflection $i = r$ follow karta hai. Prisms, lenses aur mirrors wavefronts ki shape badalte hain, aur object se image tak travel time har ray ke along same hota hai (Section 10.3.3).
Superposition: amplitudes judte hain. Do coherent sources $I = 4 I_{0} \cos^{2} (ϕ / 2)$ dete hain, path difference $n λ$ par bright aur $(n + \frac{1}{2}) λ$ par dark; incoherent sources har jagah $2 I_{0}$ dete hain (Section 10.4).
Young's experiment mein ek slit do coherent slits ko light deti hai. Bright fringes $x = n λ D / d$ par hain, dark $(n + \frac{1}{2}) λ D / d$ par, barabar doori par, fringe width $β = λ D / d$ ke saath (Section 10.5).
Width $a$ ki single slit ke liye central maximum ki angular width $2 λ / a$ hai, minima $θ = n λ / a$ par hain, aur secondary maxima kaafi weak hain. Asli double-slit pattern is envelope ke andar fringes hai (Section 10.6.1).
Diffraction yeh limit karta hai ki optical instruments kitni detail resolve kar sakte hain. Interference aur diffraction sirf energy ko redistribute karte hain (Section 10.6).
Light transverse hai aur polarise ho sakti hai. Polaroid $𝐄$ ka apne pass-axis ke along wala component guzarne deta hai: yeh unpolarised light ko aadha kar deta hai, aur polarised light Malus' law follow karti hai, $I = I_{0} \cos^{2} θ$ (Section 10.7).

Flash cards · 11

Key relations

1 / 11

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 10.1	Practice Q1
NCERT Exercises 10.2	Practice Q2
NCERT Exercises 10.3	Practice Q3
NCERT Exercises 10.4	Practice Q4
NCERT Exercises 10.5	Practice Q5
NCERT Exercises 10.6	Practice Q6

Quick check: MCQs

10.1 Introduction3 questions
10.2 Huygens principle3 questions
10.3.3 Plane surface se plane wave ka reflection4 questions
10.4 Waves ka coherent aur incoherent addition4 questions
10.5 Light waves ka interference aur Young's experiment4 questions
10.6.2 Single slit diffraction pattern dekhna4 questions
10.7 Polarisation4 questions

Glossary · 14

Key terms (English + Hindi)

coherent sourcesकला-संबद्ध स्रोतkala-sambaddh srot: Same frequency ke sources jinka phase difference constant rehta hai. Incoherent sources ka phase difference randomly badalta rehta hai.Chapter mein pehli baar yahan aaya →
critical angleक्रांतिक कोणkrantik kon: Denser medium mein woh angle of incidence jiske liye angle of refraction $90^{\circ}$ hai: $\sin i_{c} = n_{2} / n_{1}$ .Chapter mein pehli baar yahan aaya →

diffractionविवर्तनvivartan: Obstacles ke around aur openings se hokar wave ka phailna, jo tab dikhta hai jab woh wavelength se bahut bade na hon.Chapter mein pehli baar yahan aaya →

fringe widthफ्रिंज चौड़ाईfringe chaudai: Paas-paas ki bright (ya dark) fringes ke beech ki distance; double slit ke liye $β = λ D / d$ .Chapter mein pehli baar yahan aaya →

interferenceव्यतिकरणvyatikaran: Coherent waves ke overlap hone par intensity ka redistribution: jahan woh step mein pahunchti hain wahan bright, jahan half cycle ke farq se pahunchti hain wahan dark.Chapter mein pehli baar yahan aaya →

linearly polarisedरैखिकतः ध्रुवितraikhikatah dhruvit: Aisi light jiska electric field ek fixed line ke along oscillate karta hai; ise plane polarised bhi kehte hain.Chapter mein pehli baar yahan aaya →

Malus' lawमैलस का नियमMailas ka niyam: Intensity $I_{0}$ wali polarised light jab aise polaroid se milti hai jiska axis usse $θ$ par ho, to $I = I_{0} \cos^{2} θ$ .

path differenceपथांतरpathantar: Do sources se kisi point tak ki distances ka difference; isse phase difference $ϕ = 2 π Δ / λ$ milta hai.Chapter mein pehli baar yahan aaya →
polarisationध्रुवणdhruvan: Woh direction jismein transverse wave ki vibration hoti hai (light ke liye, $𝐄$ ).Chapter mein pehli baar yahan aaya →
polaroidपोलेरॉइडpoleroid: Line mein lage long-chain molecules wali plastic sheet jo $𝐄$ ka sirf apne pass-axis ke along wala component guzarne deti hai.Chapter mein pehli baar yahan aaya →

secondary waveletद्वितीयक तरंगिकाdvitiyak tarangika: Huygens' construction mein, wavefront ke kisi point se phailti ek chhoti spherical wave.Chapter mein pehli baar yahan aaya →
superposition principleअध्यारोपण का सिद्धांतadhyaropan ka siddhant: Overlap karti waves ka displacement un displacements ke vector sum ke barabar hota hai jo har wave akele paida karti.Chapter mein pehli baar yahan aaya →

unpolarised lightअध्रुवित प्रकाशadhruvit prakash: Aisi light jiske electric field ki direction ray ke perpendicular plane ke andar tezi se aur randomly badalti rehti hai.Chapter mein pehli baar yahan aaya →

wavefrontतरंगाग्रtarangagra: Aisi surface jis par wave ka har point same phase mein oscillate karta hai. Rays iske perpendicular hoti hain.Chapter mein pehli baar yahan aaya →

Wave Optics: Class 12 Physics Chapter 10 Hinglish mein

Is chapter mein aap kya seekhenge

10.1 Introduction

10.2 Huygens principle

10.3 Huygens principle se plane waves ka refraction aur reflection

10.3.1 Plane wave ka refraction

10.3.2 Rarer medium par refraction

10.3.3 Plane surface se plane wave ka reflection

10.4 Waves ka coherent aur incoherent addition

10.5 Light waves ka interference aur Young's experiment

10.6 Diffraction

10.6.1 Single slit

10.6.2 Single slit diffraction pattern dekhna

10.7 Polarisation

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye