Skip to chapter text
PhysicsCh 9 Ray Optics and Optical Instruments

Ray Optics and Optical Instruments: Class 12 Physics · Chapter 9

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 115 min
  • 36 practice questions
  • 27 MCQs

What you will learn in Ray Optics and Optical Instruments

  • Explain when light can be treated as rays and apply the Cartesian sign convention to mirrors and lenses
  • Derive f = R/2 and the mirror equation, and use them with the magnification to locate and describe images
  • Apply Snell's law, explain lateral shift and apparent depth, and find critical angles for total internal reflection
  • Explain how totally reflecting prisms and optical fibres work
  • Derive the refraction formula for a spherical surface, the lens maker's formula and the thin lens formula, and use the power of lenses alone and in contact
  • Derive the deviation produced by a prism and find the refractive index from the angle of minimum deviation
  • Describe and calculate the magnifying power of a simple microscope, a compound microscope and refracting and reflecting telescopes

Original companion text aligned to NCERT Physics Class 12, Chapter 9, pp. 221–254. Not affiliated with or endorsed by NCERT.

9.1 Introduction

Sunlight slanting through a gap in a tin roof, or a torch beam in a dusty room, draws a sharp straight line through the air. Shadows have crisp edges, and a tall building hides the Sun exactly when it blocks the straight line from your eye. Everyday experience says two things about light: it goes very fast, and it goes in straight lines. This chapter builds a whole toolkit for mirrors, lenses, prisms and instruments on the second idea.

In Chapter 8 of Part I you met light as an electromagnetic wave. Our eyes respond to a narrow slice of that spectrum, wavelengths from about 400 nm (violet) to about 750 nm (red), and we call that slice light (प्रकाश). In vacuum it travels at c=2.99792458×108 m/s, a value now fixed by definition; 3.00×108 m/s is good enough for almost every calculation. Nothing carries energy or information faster. Even so, sunlight needs about 500 s to cover the 1.5×1011 m to the Earth.

How can a wave travel in straight lines? The answer is size. A wave squeezing through an opening spreads out noticeably only when the opening is comparable to its wavelength. Visible light has λ≈5×10−7 m, while a window, a lens or a coin is a centimetre to a metre across, a hundred thousand to a million times larger. Past such objects the spreading is far too small to notice, and the energy follows straight lines (Fig. 9.1). Chapter 10 deals with what happens when openings do become tiny.

3/3Lines drawn along the direction of travel, perpendicular to the wavefronts, are rays; a bundle of them is a beam.

Fig. 9.1When openings are huge compared with the wavelength, light energy travels along straight lines, the rays.

A small lamp on the left sends out circular wavefronts. A wall with a wide window stands in the middle, and a second wall is on the right. Step 1: the wavefronts spread out from the lamp and the part that passes the window continues as arcs. Step 2: the light beyond the window is shaded as a straight-edged wedge that makes a sharply bounded bright patch on the far wall, labelled sharp edge. Step 3: three arrows from the lamp through the window to the wall are drawn as rays, perpendicular to the wavefronts, and the label says a beam is a bundle of rays.

Definition
Ray

A straight line along which light energy travels, drawn perpendicular to the wavefronts. A bundle of rays is called a beam: parallel, converging or diverging.

ExampleThe thin bright line a laser pointer draws in a misty room is a good picture of a single ray.

With the ray picture we study three things in this chapter: reflection, refraction and the images they form at plane and curved surfaces. First come spherical mirrors, then refraction at flat surfaces, total internal reflection, curved surfaces and thin lenses, and the prism. The chapter ends with the simple microscope, the compound microscope and the telescope, instruments built from these parts.

Quick check
1 of 3
The Moon is 3.84×108 m from the Earth. How long, in seconds, does moonlight take to reach us?
s
Show answer

Answer 1.28 s

t=d/c=3.84×108/3.00×108=1.28 s. A laser pulse sent to a reflector on the Moon comes back after about 2.56 s.

9.2 Reflection of light by spherical mirrors

You already know the two laws of reflection. The angle of reflection equals the angle of incidence, both measured from the normal; and the incident ray, the normal and the reflected ray lie in one plane. Nothing in these laws says the surface must be flat. At any point of a curved mirror, a tiny patch around that point is effectively flat, so the laws apply point by point, with the normal to that patch.

For a spherical mirror, a piece cut from a hollow sphere, the normal at every point is easy to find: it lies along the radius through that point, so every normal passes through the centre of curvature C. The middle point of the mirror's surface is its pole P, the distance PC is the radius of curvature R, and the line through P and C is the principal axis. A mirror silvered on its hollow side is concave; one silvered on its bulging side is convex. For a lens the corresponding middle point is the optical centre, and its principal axis is the line through the optical centre and the principal foci.

One more requirement hides in the word image. A polished mirror reflects regularly: neighbouring rays meet neighbouring patches that face the same way, so rays from one object point stay together. A whitewashed wall reflects most of the light that falls on it, yet you cannot see your face in it: its rough grains face every direction, so rays from one point scatter everywhere and never come together at one image point.

9.2.1 Sign convention

A mirror formula has to cover objects in front of a mirror and images behind it, concave and convex mirrors, real and virtual images. Instead of a separate formula for every case, we give every distance a sign and use one formula throughout. This book uses the Cartesian sign convention (Fig. 9.2):

  1. Origin. Measure from the middle of the mirror or lens, its pole P (for a lens, its optical centre O), with the principal axis as the x-axis.
  2. Distances. A point reached by moving with the incident light from the origin has a positive distance; one reached by moving against the light has a negative distance.
  3. Heights. Points above the axis have positive heights, points below it negative heights.

4/4Upward heights are positive, downward heights negative, so an inverted image has a negative height.

Fig. 9.2Measure from the pole; the direction of the incident light is positive.NCERT Fig. 9.2 · p. 222

Light comes in from the left towards a convex mirror whose pole P is in the middle of the drawing. Step 1: arrows from P mark distances to the right, along the incident light, as positive and distances to the left as negative. Step 2: an object stands on the axis to the left; its distance from P is drawn as an arrow pointing left and labelled u less than zero. Step 3: a small upright dashed image appears behind the mirror, and the focus F lies further behind; both distances point right and are labelled v greater than zero and f greater than zero. Step 4: a key shows heights upward as positive and downward as negative, and the object height is labelled h greater than zero.

Definition
Cartesian sign convention

Distances are measured from the pole (mirror) or optical centre (lens): positive in the direction of the incident light, negative against it. Heights above the principal axis are positive, below it negative.

ExampleWith light from the left, a real object 30 cm in front of a mirror has u=−30 cm; a concave mirror of focal length 15 cm has f=−15 cm.

We always draw the incident light travelling from left to right. Then a real object is on the left and has a negative u; the focus of a concave mirror is in front of it, so its f is negative; the focus of a convex mirror is behind it, so its f is positive. Do not decide signs by looking at whether an image is real. Put the numbers with their signs into the formula, and let the sign of the answer tell you where the image is.

9.2.2 Focal length of spherical mirrors

Send a narrow beam parallel to the principal axis onto a concave mirror. If the rays are paraxial (they hit the mirror near its pole and are only slightly inclined to the axis), they all reflect through a single point F on the axis: the principal focus. A convex mirror spreads such a beam out, and the reflected rays appear to come from a point F behind it (Fig. 9.3). The distance PF is the focal length f.

A parallel beam that is tilted to the axis also comes to a point, but off the axis. All such points lie in the plane through F perpendicular to the axis, the focal plane. This is why a distant scene, whose points send parallel beams in slightly different directions, forms a flat image near the focus.

1/3Paraxial rays parallel to the axis reflect through the principal focus F, in front of the mirror.

Fig. 9.3A concave mirror brings a parallel paraxial beam to a focus; a convex mirror makes it diverge from one.NCERT Fig. 9.3 · p. 223

The principal axis runs across the figure. Step 1: four rays parallel to the axis strike a concave mirror on the right and all reflect through the point F, halfway between the pole P and the centre of curvature C; the focal length f is marked from F to P. Step 2: the same rays strike a convex mirror and reflect outwards; dashed backward extensions of the reflected rays meet at F behind the mirror. Step 3: a parallel beam arrives at an angle to the axis and, after reflection from the concave mirror, meets at a point below the axis on a dashed vertical line through F, labelled focal plane.

DerivationWhy the focal length is half the radius

Take a ray parallel to the axis that strikes a concave mirror at Q (Fig. 9.4). The radius CQ is the normal at Q; call the angle of incidence θ.

Fig. 9.4The normal CQ makes angle θ with the axis at C, and the reflected ray makes θ with the normal at Q: triangle CFQ has two equal angles.NCERT Fig. 9.4 · p. 224

A concave mirror arc on the right with centre of curvature C on the axis at the left. A ray parallel to the axis strikes the mirror at Q. The dashed normal CQ, of length R, makes angle theta with the axis at C. At Q the incident ray and the reflected ray each make angle theta with the normal. The reflected ray crosses the axis at F, where it makes angle 2 theta with the axis, and continues to the lower left. The pole P is where the axis meets the mirror.

The incident ray is parallel to the axis, so the normal CQ meets the axis at the same angle θ (alternate angles): ∠FCQ=θ. The law of reflection makes the reflected ray QF lie at θ on the other side of the normal: ∠CQF=θ. Triangle CFQ therefore has two equal angles, so it is isosceles:

CF=FQ

(The exterior angle at F is 2θ, which is why the reflected ray crosses the axis at twice the angle of incidence.) Now use the paraxial condition. When Q is close to the pole, the slanting length FQ is practically the same as FP, so CF≈FP: the focus sits halfway between C and P. Since CP=R and FP=f,

focal length of a spherical mirror
f=R2
(9.3)

The same argument for a convex mirror puts F halfway between P and C, behind the mirror. With signs, a concave mirror has R and f both negative and a convex mirror has both positive, and f=R/2 holds either way. The argument also shows the price of the approximation. For a ray far from the axis, FQ is noticeably longer than FP, so F creeps towards the mirror and a wide spherical mirror gives a blurred focus. Paraxial rays avoid this; torch reflectors avoid it by using a parabolic shape.

9.2.3 The mirror equation

Light from a point on an object spreads in every direction. If, after reflection, all those rays actually pass through one point, that point is a real image (वास्तविक प्रतिबिंब) of the object point; if they only seem to diverge from a point when extended backwards, it is a virtual image (आभासी प्रतिबिंब). An image is a point-to-point copy of the object made by reflection or refraction.

To find an image point we need only two rays, because two lines fix their crossing point. Four rays from a concave mirror are easy to draw (Fig. 9.5):

  • a ray parallel to the axis reflects through F (for a convex mirror, it reflects as if coming from F);
  • a ray through F (or heading for F of a convex mirror) reflects parallel to the axis;
  • a ray through C (or heading for C of a convex mirror) meets the mirror along a normal and comes straight back;
  • a ray striking the pole reflects at an equal angle on the other side of the axis.

5/5A ray striking P reflects symmetrically and also passes through the image: here it is twice the size of the object.

Fig. 9.5Any two of the four easy rays fix the image; the others pass through the same point.NCERT Fig. 9.5 · p. 224

A concave mirror on the right with its pole P, focus F and centre of curvature C on the axis, and an upright object arrow between F and C. Step 1 shows only this set-up. Step 2: a blue ray from the top of the object travels parallel to the axis, reflects and passes through F. Step 3: an orange ray from the object top passes through F, hits the mirror and returns parallel to the axis; where it crosses the blue ray, beyond C, an inverted image arrow appears. Step 4: a green ray along the line from C through the object top strikes the mirror and comes straight back through C to the same image point. Step 5: a purple ray strikes the pole and reflects at an equal angle below the axis, also passing through the image top. The note reads real, inverted, twice as tall.

The diagram shows only a few rays, but the object point sends out countless rays. Every one that reaches the mirror passes through the same image point, which is what makes the image sharp. Now we turn the drawing into an equation.

DerivationThe mirror equation

Place the pole at the origin with the axis as the x-axis and light travelling in the +x direction. Let the object tip be at (u,h), the image tip at (v,h′) and the centre of curvature at (R,0), all with their signs. We use two of the four rays from Fig. 9.5.

The ray to the pole. At P the normal is the axis itself, so reflection reverses the ray's horizontal direction but not its vertical one: the reflected ray keeps descending with the same steepness h/∣u∣. The incident ray reaches the axis from height h at x=u; the reflected ray, at x=v, has height h′=−hv/u. So

h′h=−vu

The ray through C. It travels along a radius, hits the mirror normally and retraces its path, so the object tip, C and the image tip lie on one straight line through (R,0). Heights along a straight line through C grow in proportion to the distance from C:

h′h=v−Ru−R

Both describe the same image, so −vu=v−Ru−R. Cross-multiplying, −uv+vR=uv−uR, or uR+vR=2uv. Divide by uvR and use R=2f:

mirror equation
1v+1u=2R=1f
(9.7)
where
u
object distance
v
image distance
f
focal length, R/2

No step needed to know whether the image was real or virtual, or which kind of mirror we had: the signs carry that information. So (9.7) holds for concave and convex mirrors, and for real and virtual images, provided every quantity goes in with its sign.

How tall the image is, compared with the object, is measured by the linear magnification:

magnification
m=h′h
(9.8)
magnification of a mirror
m=−vu
(9.9)

For a real image u and v are both negative, so m is negative: real images formed by a single mirror are inverted. For a virtual image v is positive, m is positive and the image is erect. The size of ∣m∣ says whether the image is enlarged (∣m∣>1) or diminished. Fig. 9.6 shows the two everyday virtual images: a concave mirror with the object inside F (a shaving or make-up mirror) and a convex mirror (the mirror on an auto-rickshaw).

Fig. 9.6Both images are behind the mirror and upright; the concave one is enlarged, the convex one diminished.NCERT Fig. 9.6 · p. 226

Two panels. Top: a concave mirror with an object between its focus F and the pole P. A ray parallel to the axis reflects through F, and a ray coming from the direction of F reflects parallel; their dashed backward extensions meet behind the mirror at the top of an enlarged, upright, dashed image labelled virtual. Bottom: a convex mirror with an object far in front. A ray parallel to the axis reflects as if from the focus F behind the mirror, and a ray aimed at the centre of curvature comes straight back; their dashed extensions meet behind the mirror, between the pole and F, at the top of a small upright dashed image.

InteractiveMirror bench

An interactive ray diagram. Two buttons choose a concave or a convex mirror. Sliders set the object distance from 4 to 60 centimetres and the focal length from 5 to 25 centimetres. The drawing shows the mirror, its focus and centre of curvature, an object 2 centimetres tall, the parallel ray, the ray to the pole and the ray along a radius, with dashed extensions when the image is virtual, and the image itself. Readouts give the signed object distance, image distance, focal length, magnification and image height, and a status line describes the image as real or virtual, inverted or erect, enlarged or diminished.

Fig. 9.7Choose a concave or convex mirror, then move the object and change the focal length. Read the signed values of u, v, f and m.

Try the mirror bench: start with a concave mirror and slide the object from far away towards the mirror. The real image grows and moves outwards, races off to infinity as the object reaches F, and reappears behind the mirror as a large virtual image. A convex mirror never does anything dramatic: its image is always virtual, upright and smaller.

Example 9.1NCERT Example 9.1 · p. 226

A paper disc on a shaving mirror

A concave shaving mirror forms a sharp real image of a lamp on a wall. A circular paper disc whose radius is half the mirror's radius is stuck over the centre of the mirror. What happens to the image on the wall?

Try it yourself first. Then reveal the solution one step at a time.

Example 9.2NCERT Example 9.2 · p. 226

A pencil lying along the axis

A 5.0 cm pencil lies along the axis of a concave mirror (f=10 cm), with its tip 25 cm and its blunt end 30 cm from the mirror. Find the length of its image and explain why the image is distorted.

Try it yourself first. Then reveal the solution one step at a time.

Fig. 9.8A 5 cm pencil between 25 cm and 30 cm images to a 1.7 cm sliver between 15 cm and 16.7 cm.NCERT Fig. 9.7 · p. 227

A concave mirror on the right with F at 10 centimetres and C at 20 centimetres from it. A pencil 5 centimetres long lies on the axis between 25 and 30 centimetres from the mirror, its sharp tip pointing towards the mirror. Its image lies between C and F, only 1.7 centimetres long, thinner, and with its tip pointing away from the mirror.

Example 9.3NCERT Example 9.3 · p. 227

Two positions in front of a concave mirror

A concave mirror has a radius of curvature of 24 cm. An object is placed (a) 18 cm and (b) 8.0 cm in front of it. Locate the image each time, say whether it is real or virtual, and find m.

Try it yourself first. Then reveal the solution one step at a time.

Example 9.4NCERT Example 9.4 · p. 227

How fast does an image move in a rear-view mirror?

An auto-rickshaw is parked with its convex rear-view mirror (R=1.6 m) facing a scooter that approaches at a steady 10 m/s. How fast does the scooter's image move when the scooter is (a) 50 m, (b) 10 m and (c) 2.0 m from the mirror?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A convex mirror has a radius of curvature of 40 cm. With the Cartesian sign convention, its focal length is
Show answer

Answer (C) +20 cm

f=R/2=20 cm in size, and the focus of a convex mirror lies behind it, in the direction of the incident light, so f=+20 cm. −20 cm would be a concave mirror.

9.3 Refraction

Put a straw in a glass of nimbu-pani and it looks snapped at the surface. When light meets the boundary between two transparent media, part of it reflects and the rest crosses into the second medium. If it arrives obliquely (0∘<i<90∘), the transmitted light changes direction at the boundary. This change of direction is refraction (अपवर्तन), and its rules were found by measurement long before anyone could explain them (Fig. 9.9).

LawSnell's laws of refraction

(i) The refracted ray stays in the plane defined by the incident ray and the normal at the point where it strikes. (ii) For a given pair of media and a given colour of light, sin⁡i/sin⁡r takes the same value at every angle of incidence.

Snell's law
sin⁡isin⁡r=n21
(9.10)
where
i
angle of incidence, from the normal
r
angle of refraction, from the normal
n21
refractive index of medium 2 relative to medium 1

The constant n21 is the refractive index (अपवर्तनांक) of medium 2 relative to medium 1. It belongs to the pair of media and changes slightly with colour (wavelength), but it does not depend on the angle. If n21>1, then r<i: the ray bends towards the normal, and medium 2 is called optically denser. If n21<1, the ray bends away from the normal on entering the optically rarer medium.

2/3Most light enters the water, bent towards the normal (r<i); a little is reflected.

Fig. 9.9Into the denser water the ray bends towards the normal; reversed, it bends away along the same path.NCERT Fig. 9.8 · p. 228

Air above, water below, separated by a horizontal boundary with a dashed vertical normal. Step 1: a ray in air arrives at the boundary at angle i from the normal. Step 2: in the water the ray continues at a smaller angle r, bent towards the normal, and a weak dashed ray is reflected back into the air. Step 3: the arrows are reversed: a ray coming up from the water at angle i refracts into the air at the larger angle r, bending away from the normal along the same path.

Refraction happens because light travels at different speeds in different media. The absolute refractive index of a medium is n=c/v, where v is the speed of light in it; for vacuum n=1 and for air n=1.0003, which we take as 1. In terms of absolute indices, n21=n2/n1=v1/v2, and Snell's law takes the symmetric form n1sin⁡i=n2sin⁡r. Water (n=1.33) slows light to 2.26×108 m/s; diamond (n=2.42) slows it to 1.24×108 m/s.

Because a light path can always be run backwards (Fig. 9.9, step 3), the index of medium 1 relative to medium 2 is just the reciprocal:

n12=1n21
(9.11)

Relative indices also chain together. Since n32=n3/n2=(n3/n1)(n1/n2), we have n32=n31×n12. For example, glass relative to water is nglass, water=1.50/1.33=1.13: a ray going from water into glass bends only a little.

A parallel-sided slab. A ray crossing a glass block refracts twice (Fig. 9.10). The two faces are parallel, so the normals are parallel and the ray meets the second face at the same angle r at which it left the first. Snell's law at the exit, nsin⁡r=sin⁡e, then gives e=i. The ray leaves in the same direction it arrived: there is no deviation, only a sideways lateral shift d. From the geometry of the figure, d=tsin⁡(i−r)/cos⁡r for a slab of thickness t. A 6.0 cm slab with n=1.50 and i=50∘ gives r=30.7∘ and d=2.3 cm.

Fig. 9.10The ray leaves the slab in its original direction, displaced sideways by d.NCERT Fig. 9.9 · p. 229

A rectangular glass slab of thickness t with air above and below. A ray enters the top face at angle i, travels through the glass at the smaller angle r, and leaves the bottom face at angle e equal to i. A dashed line continues the original ray straight through the glass; the emergent ray runs parallel to it, displaced by the perpendicular distance d, which is marked with a double-headed arrow.

Apparent depth. A coin at the bottom of a bucket of water looks nearer the surface than it is. Rays from the coin bend away from the normal as they leave the water, so when your eye traces them back in straight lines they seem to come from a point higher up (Fig. 9.11).

3/3Traced back in straight lines (dashed), they meet above the coin: the eye sees it at the apparent depth.

Fig. 9.11Seen from above, the coin appears raised: apparent depth = real depth / n.NCERT Fig. 9.10 · p. 229

A tumbler of water seen from the side with a coin on the bottom and an eye above. Step 1: two rays leave the coin at small angles to the vertical and reach the surface. Step 2: at the surface they bend away from the normal as they enter the air and travel on to the eye. Step 3: dashed lines extend the rays in air backwards into the water, where they meet at a point above the coin, marked image. A double arrow on the left shows the real depth from the surface to the coin; a shorter one on the right shows the apparent depth to the image.

DerivationApparent depth for near-normal viewing

A ray leaves the coin, at real depth hr, at a small angle θ to the vertical and reaches the surface a horizontal distance x=hrtan⁡θ from the point above the coin. In air it travels at angle θ′, with sin⁡θ′=nsin⁡θ. Traced back, it crosses the vertical through the coin at a depth ha with x=hatan⁡θ′. Equating the two expressions for x and using tan⁡≈sin⁡ for small angles:

ha=hr tan⁡θtan⁡θ′≈hr sin⁡θsin⁡θ′=hrn

The result does not depend on θ, so all near-vertical rays seem to come from one point: a clear image. A swimming pool 2.0 m deep looks only 1.5 m deep from the edge, a well-known trap for non-swimmers. Viewed obliquely, the rays no longer agree on one point and the image rises even further.

Quick check
1 of 4
A ray in air strikes glass (n=1.50) at 40∘ to the normal. Find the angle of refraction in degrees.
°
Show answer

Answer 25.4 °

sin⁡r=sin⁡40∘/1.50=0.6428/1.50=0.4285, so r=25.4∘. The ray bends towards the normal.

9.4 Total internal reflection

Now send light the other way, from a denser medium into a rarer one, say from water into air. Some light is reflected back into the water at the surface, which is called internal reflection, and the rest refracts out, bending away from the normal so that r>i. As i grows, r grows faster, and at one particular angle of incidence the refracted ray skims along the surface with r=90∘. That angle is the critical angle ic (क्रांतिक कोण) (Fig. 9.12).

Beyond ic, Snell's law would need sin⁡r>1, which is impossible: there is no refracted ray at all, and all the light is reflected back into the denser medium. This is total internal reflection (पूर्ण आंतरिक परावर्तन). Even the best silvered mirror absorbs a few per cent of the light that falls on it; total internal reflection loses essentially none, which is why it is so useful.

4/4At 62∘ nothing gets out: the light is totally internally reflected.

Fig. 9.12As the angle of incidence passes the critical angle, refraction stops and all the light is reflected.NCERT Fig. 9.11 · p. 230

A lamp sits under water with air above. Each step shows one ray from the lamp to the surface with the normal drawn at the point where it arrives; earlier rays remain as faint grey lines. Step 1: at 25 degrees the ray refracts into the air at 34 degrees and a weak dashed ray reflects. Step 2: at 40 degrees the refracted ray bends further from the normal, to 59 degrees. Step 3: at the critical angle, about 48.8 degrees, the refracted ray grazes along the surface. Step 4: at 62 degrees there is no refracted ray and the whole ray is reflected back down into the water.

Setting r=90∘ in Snell's law for light going from medium 1 (denser) into medium 2 (rarer), n1sin⁡ic=n2:

critical angle
sin⁡ic=n2n1=n21
(9.12)

For a material of index n in air this is sin⁡ic=1/n, so measuring the critical angle gives the refractive index: n=1/sin⁡ic. The larger the index, the smaller the critical angle (Table 9.2).

Table 9.2Critical angles for light leaving some transparent media into air (sin⁡ic=1/n)NCERT Table 9.1 · p. 230
MediumnCritical angle
Ice1.3149.8°
Water1.3348.8°
Acrylic plastic1.4942.2°
Crown glass1.5241.1°
Dense flint glass1.6238.1°
Diamond2.4224.4°
InteractiveRefraction lab

An interactive diagram of a ray meeting a flat boundary. Six buttons choose the pair of media: air to water, air to glass, water to air, glass to air, glass to water and diamond to air. A slider sets the angle of incidence from 0 to 89 degrees. The drawing shows the incident ray, the reflected ray, whose thickness grows with the fraction of light reflected, and the refracted ray when there is one. Readouts give both refractive indices, the angles of incidence and refraction, the products n sin i and n sin r, the critical angle and the reflected percentage; a status line explains whether the ray bends towards or away from the normal or is totally internally reflected.

Fig. 9.13Pick two media and change the angle of incidence. Watch the refracted ray, the share of light reflected and, going into a rarer medium, the critical angle.

9.4.1 Total internal reflection in nature and its technological applications

(i) Prisms that act as perfect mirrors. A glass prism whose angles are 45∘, 45∘ and 90∘ can turn light without any silvering. Light entering one face normally meets the long face at 45∘, more than the critical angle of glass (41–42∘, Table 9.2), and is totally reflected. Depending on how the prism is turned, it bends the beam through 90∘, sends it back through 180∘, or flips an image upside down without changing its size (Fig. 9.14). Periscopes and binoculars use such prisms; their reflections stay bright for decades because there is no coating to tarnish.

2/3Two total reflections return the light the way it came, as in a bicycle reflector or the prism of binoculars.

Fig. 9.14Total internal reflection at 45° makes a prism a loss-free mirror.NCERT Fig. 9.13 · p. 231

Step 1: two rays enter the vertical face of a right-angled prism, meet the sloping long face at 45 degrees, are totally reflected and leave downwards through the bottom face, turned through 90 degrees. Step 2: the prism is turned so that the long face faces the light; two rays enter it, are totally reflected at each short face in turn and come back parallel to their original direction, with their order swapped. Step 3: the prism lies on its long face; rays A and B enter the left sloping face one above the other, refract, are totally reflected at the base and leave the right sloping face with B now above A.

(ii) Optical fibres. An optical fibre (प्रकाशिक तंतु) is a glass or plastic thread thinner than a hair. Its core has a slightly higher refractive index than the surrounding cladding. Light entering one end at a suitable angle strikes the core–cladding boundary at more than the critical angle, is totally reflected, crosses to the other side, and is reflected again, thousands of times per metre (Fig. 9.15). A fibre can even be bent gently: the light follows it round the bend, which is why it is often called a light pipe.

Fig. 9.15Each time the ray meets the core–cladding boundary it is totally reflected, so the light stays in the core.NCERT Fig. 9.14 · p. 232

A long straight fibre drawn in cross-section: a central core of higher refractive index between two layers of cladding of lower refractive index. A ray enters the left end, crosses the core, strikes the core-cladding boundary at a glancing angle, is totally reflected, and continues to zig-zag along the core until it leaves the right end.

Fibres carry telephone calls, television and internet data as flashes of light, both under the oceans and into homes. Bundles of fibres also carry light into the body and an image back out: with such an endoscope doctors can look inside the stomach, the airways or a damaged knee without large cuts. The tufted lamps sold at fairs, whose plastic strands glow at their tips, are decorative optical fibres lit from a bulb at the base.

The real challenge is purity. Ordinary window glass looks clear, but a block a few metres thick would be dark green. Silica for fibres is purified so carefully that more than 95% of the light survives a full kilometre.

Example

How steeply can light enter a fibre?

A fibre has a core of index 1.50 and a cladding of index 1.46. (a) What is the critical angle at the core–cladding boundary? (b) What is the largest angle a ray inside the core can make with the fibre's axis and still be guided? (c) What is the largest angle to the axis at which light from air can enter the flat end and be guided?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
Find the critical angle for light going from glass (n=1.50) into water (n=1.33), in degrees.
°
Show answer

Answer 62.5 °

sin⁡ic=n2/n1=1.33/1.50=0.887, so ic=62.5∘. It is larger than the glass–air value of 41.8∘ because the two indices are closer.

9.5 Refraction at spherical surfaces and by lenses

A drop of water on a leaf, a glass marble and the lens of your eye all refract light at curved surfaces. As with curved mirrors, we treat a tiny patch of a spherical surface as flat, apply Snell's law there, and remember that the normal at any point runs along the radius, through the centre of curvature. We first find what a single spherical surface does. A thin lens is then just two such surfaces, close together, at least one of them curved: applying the single-surface result twice gives the lens maker's formula and the lens formula.

9.5.1 Refraction at a spherical surface

Medium 1 (index n1, on the left) meets medium 2 (index n2, on the right) at a convex spherical boundary whose centre is C and whose radius is R. A point object O sits on the axis in the first medium, and P is where the axis meets the surface. We want the point I where the refracted rays cross the axis. Keep the aperture small compared with all other distances, so every angle in Fig. 9.16 is small.

DerivationImage formed by one refracting surface
Fig. 9.16The angles at O, C and I are α, β and γ; at A, i=α+β and r=β−γ.NCERT Fig. 9.15 · p. 233

The principal axis runs left to right through the object point O, the point P where the axis meets the surface, the centre of curvature C and the image point I. The surface curves through A, a point above the axis; the region to the right of it is the second medium. A ray from O to A makes angle alpha with the axis; the radius CA, drawn dashed and extended outside, is the normal at A and makes angle beta with the axis at C. Inside, the refracted ray from A to I makes angle gamma with the axis. At A the angle of incidence i is measured between the extended incident ray and the normal, and the angle of refraction r between the refracted ray and the normal.

Let the ray from O meet the surface at A, a small height h above the axis, and call the angles that OA, CA and AI make with the axis α, β and γ. Each angle is small, so it is practically h divided by the horizontal distance from P:

α≈hOP,β≈hPC,γ≈hPI

Treat O, A and C as a triangle: the angle of incidence i (between the incoming ray, extended, and the normal) is its exterior angle at A, so i=α+β. Likewise in triangle ACI the angle β at C is exterior, so β=r+γ, or r=β−γ:

i=hOP+hPC,r=hPC−hPI
(9.14)

Snell's law for small angles is n1i=n2r. Substituting and cancelling h,

n1OP+n2PI=n2−n1PC
(9.15)

These are lengths. With the Cartesian convention, measured from P along the incident light, OP=−u, PI=+v and PC=+R:

refraction at a spherical surface
n2v−n1u=n2−n1R
(9.16)
where
n1
index of the medium the light comes from
n2
index of the medium it enters
R
radius of curvature, positive if C is on the outgoing side

With signs, (9.16) holds for convex and concave surfaces and for real and virtual images alike. For a flat surface R→∞ and it gives v=(n2/n1)u, the apparent-depth result of Section 9.3 once more.

Example 9.5NCERT Example 9.5 · p. 233

Images at a curved glass surface

(a) A point source in air is 40 cm from the convex surface of a long glass rod (n=1.50) whose end is ground to a radius of 10 cm. Where is the image formed? (b) A small air bubble sits exactly at the centre of a solid glass marble. Where does it appear to someone looking at the marble from outside?

Try it yourself first. Then reveal the solution one step at a time.

9.5.2 Refraction by a lens

A thin lens of index n2 in a medium of index n1 has two surfaces with radii R1 (the first surface the light meets) and R2. The trick is to let the first surface act alone, as if glass filled all the space beyond it, and find the image I1 it would form. That image then serves as the object for the second surface (Fig. 9.17).

3/3Surface 2 bends the rays again and they meet at the final image I.

Fig. 9.17The first surface would form I1; the second surface takes I1 as its object and forms the final image I.NCERT Fig. 9.16 · p. 234

A point object O on the left of a double convex lens. Step 1: only the first surface is drawn, with glass filling all the space to its right; a ray from O refracts at this surface and heads for the point I1 far to the right. Step 2: the whole lens is drawn; the ray inside the glass still heads for I1 (dashed beyond the lens), so I1 is a virtual object for the second surface. Step 3: at the second surface the ray refracts again and crosses the axis at the final image I, much closer to the lens.

DerivationThe lens maker's formula and the thin lens formula

First surface (from n1 into n2, radius R1), object at distance u, image I1 at v1. By (9.16):

n2v1−n1u=n2−n1R1
(9.17)

Second surface (from n2 back into n1, radius R2). Its object is I1. For a thin lens both surfaces are practically at the optical centre, so the object distance is again v1; the final image is at v. Applying (9.16) with the media swapped:

n1v−n2v1=n1−n2R2
(9.18)

Adding the two equations removes v1:

n1v−n1u=(n2−n1)(1R1−1R2)
(9.19)

An object at infinity (u→−∞) images at the second principal focus, v=f. Dividing by n1 and writing n21=n2/n1:

lens maker's formula
1f=(n21−1)(1R1−1R2)
(9.21)

The right-hand side of (9.19) divided by n1 is therefore 1/f, and

thin lens formula
1v−1u=1f
(9.23)

For a double convex lens R1>0 and R2<0, so f>0: it converges light. For a double concave lens the signs flip (R1<0, R2>0) and f<0: it diverges light. Neither formula needed to know which case we were in.

The lens maker's formula (9.21) is what a lens designer uses: pick a glass, then choose radii that give the focal length wanted. A lens has two principal foci, F on the side of the incoming light (the first focal point) and F′ on the other side (the second focal point). When the same medium is on both sides of a thin lens, the two are at the same distance ∣f∣ from the optical centre O.

For drawing, three rays through a thin lens are easy (Fig. 9.18):

  • a ray parallel to the axis passes through the second focus F′ of a convex lens, or diverges as if from the first focus F of a concave lens;
  • a ray through the optical centre O goes straight on, undeviated;
  • a ray through F of a convex lens (or heading for F′ of a concave lens) emerges parallel to the axis.

4/5A ray through F leaves parallel; the three meet: real, inverted, enlarged image beyond 2F′.

Fig. 9.18Three easy rays fix the image of a lens; for a concave lens their backward extensions meet.NCERT Fig. 9.17 · p. 235

Step 1: a convex lens with its optical centre O, foci F and F prime and points 2F and 2F prime marked on the axis; an upright object stands between 2F and F. Step 2: a blue ray from the object top travels parallel to the axis and after the lens passes through F prime. Step 3: a green ray passes straight through O. Step 4: an orange ray through F emerges parallel to the axis; all three rays meet beyond 2F prime, where an inverted, enlarged image appears; the note says real, inverted, larger. Step 5: a concave lens instead; a parallel ray diverges as if from F on the object side (dashed back to F), and the ray through the centre goes straight on; traced back, they meet between F and the lens at a small, upright, dashed image.

The magnification of a lens is defined exactly as for a mirror, and similar triangles through O give

magnification of a lens
m=h′h=vu
(9.24)

For a real image u<0 and v>0, so m<0: the image is inverted. A virtual image from a lens lies on the object side, v<0, so m>0 and the image is erect. Try both lenses on the lens bench below and check 1/v−1/u=1/f with the signed readouts.

InteractiveLens bench

An interactive ray diagram. Two buttons choose a convex or a concave lens. Sliders set the object distance from 4 to 60 centimetres and the focal length from 5 to 25 centimetres. The drawing shows the lens, its two foci and the points at twice the focal length, an object 2 centimetres tall, the parallel ray, the ray through the centre and the ray through a focus, with dashed backward extensions when the image is virtual, and the image itself. Readouts give the signed object distance, image distance, focal length, magnification and image height; a status line describes the image.

Fig. 9.19Choose a convex or concave lens, then move the object and change the focal length. Read the signed values of u, v, f and m.
Example 9.6NCERT Example 9.6 · p. 236

One lens, three liquids

A thin double convex lens is made of glass with n=1.52. It is dipped in turn into three liquids, of index (a) 1.33, (b) 1.52 and (c) 1.63. How does it behave in each, compared with its behaviour in air?

Try it yourself first. Then reveal the solution one step at a time.

9.5.3 Power of a lens

How strongly a lens bends light is measured by its power. A ray parallel to the axis at height h is bent by an angle δ so that it crosses the axis at the focus: tan⁡δ=h/f. For a ray at unit height and small angles, δ≈1/f: the shorter the focal length, the more sharply the lens converges (or, for negative f, diverges) a beam. So we define the power of a lens (लेंस की क्षमता) as

power of a lens
P=1f
(9.25)
Definition
Power of a lens

The reciprocal of the focal length in metres, P=1/f. Its SI unit is the dioptre (D): 1 D=1 m−1. Converging lenses have positive power, diverging lenses negative power.

ExampleA lens of focal length +0.25 m has a power of +4 D.

Opticians write prescriptions in dioptres. Reading glasses of +2.0 D are converging lenses with f=+50 cm; glasses for a short-sighted person of −2.5 D are diverging lenses with f=−40 cm.

Example 9.7NCERT Example 9.7 · p. 236

Power, refractive index and a lens under water

(a) What is the power of a diverging lens with f=−25 cm? (b) A double convex lens has faces of radii 12 cm and 18 cm and a focal length of 12 cm in air. Find the refractive index of its glass. (c) A glass lens (n=1.60) has a focal length of 15 cm in air. What is its focal length in water (n=1.33)?

Try it yourself first. Then reveal the solution one step at a time.

9.5.4 Combination of thin lenses in contact

Camera and microscope objectives are made of several lenses cemented together. Two thin lenses of focal lengths f1 and f2 in contact share practically the same optical centre. What single lens would do the same job?

DerivationFocal length of lenses in contact

Let the object be at u. The first lens alone would form an image at v1:

1v1−1u=1f1
(9.27)

The second lens receives the light converging towards (or diverging from) that point, so v1 is its object distance. It forms the final image at v:

1v−1v1=1f2
(9.28)

The first image is only a step in the calculation; the light never actually reaches it, because the second lens is in the way. Adding the two equations,

1v−1u=1f1+1f2
(9.29)

This is the lens formula for a single lens of focal length f with

lenses in contact
1f=1f1+1f2
(9.30)

Repeating the argument for more lenses adds more terms:

1f=1f1+1f2+1f3+⋯
(9.31)
powers add
P=P1+P2+P3+⋯
(9.32)

The sum is algebraic: diverging lenses contribute negative powers. And since each image is the next lens's object, the overall magnification is the product of the separate ones:

m=m1m2m3⋯
(9.33)

Combining lenses lets a designer reach a power that no single lens has, and, by pairing glasses that spread colours differently, cancel defects such as coloured fringes. Lens systems of this kind sit inside every camera, microscope and telescope.

Example

Two lenses in contact

A converging lens of power +4.0 D is placed in contact with a diverging lens of power −1.5 D. Find the power and focal length of the combination, and say whether it converges or diverges light.

Try it yourself first. Then reveal the solution one step at a time.

Example 9.8NCERT Example 9.8 · p. 238

Two separated lenses

An object stands 30 cm in front of a converging lens of focal length 20 cm. A diverging lens of focal length 40 cm is placed 30 cm behind the first lens, on the same axis. Find the position, nature and size of the final image relative to the object.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A double convex lens has radii 20 cm and 30 cm and is made of glass with n=1.50. What is its focal length in air, in cm?
cm
Show answer

Answer 24 cm

1/f=0.50 (1/20+1/30)=0.50×5/60=1/24, so f=24 cm. With signs, R1=+20 and R2=−30, and −1/R2 adds.

9.6 Refraction through a prism

A triangular glass prism has two refracting faces that meet at the angle of the prism A. A ray enters the first face at an angle of incidence i and refracts to r1, crosses the glass, meets the second face at r2 and leaves at the angle of emergence e. Unlike a slab, the faces are not parallel, so the two bendings add up instead of cancelling: the ray comes out turned towards the base by the angle of deviation δ (Fig. 9.20).

3/4The angle between the incident direction and the emergent ray is the deviation δ.

Fig. 9.20Both refractions turn the ray towards the base; the total turn is the deviation δ.NCERT Fig. 9.21 · p. 239

An equilateral glass prism with apex angle A at the top. Step 1: a ray from the left meets the left face at angle i to the dashed normal and refracts into the glass at the smaller angle r1. Step 2: the ray crosses the prism, meets the right face at angle r2 to its normal and leaves at the larger angle e, bending away from the normal and towards the base. Step 3: the incident ray is continued straight as a dashed line, and the emergent ray is traced back; the angle between them, delta, is the deviation. Step 4: a second ray in orange enters at a larger angle so that it crosses the prism parallel to the base and leaves at the same angle as it entered; this is the path of minimum deviation.

DerivationAngles in a prism

Call the points where the ray enters and leaves the glass E and X, and draw the normals there; they meet at a point O inside the prism. Each normal is perpendicular to its face, so the two normals cross at the same angle as the two faces: the exterior angle of triangle EOX at O equals A. The interior angles of that triangle at E and X are r1 and r2, and an exterior angle of a triangle equals the sum of the two opposite interior angles. Hence

r1+r2=A
(9.34)

At the first face the ray turns through i−r1; at the second, in the same sense, through e−r2. The total deviation is the sum:

deviation by a prism
δ=(i−r1)+(e−r2)=i+e−A
(9.35)

The deviation therefore depends on how the ray enters. As i increases from small values, δ first falls, reaches a minimum, and then rises again. Explore the curve with the prism lab below. Apart from the very bottom of the curve, every deviation occurs for two angles of incidence. That is no accident: (9.35) is unchanged if i and e swap, and swapping them is just running the light backwards along the same path, which gives the same deviation.

InteractivePrism lab

An interactive prism. The upper drawing shows a ray entering the left face of a prism and leaving the right face, with dashed normals and a dashed continuation of the incident ray; if the ray is totally reflected at the second face, or reaches the base, the drawing and a message say so. Below it, a graph plots the deviation against the angle of incidence from 0 to 90 degrees for the current prism, with a dot for the present ray and a dashed line at the minimum deviation. Sliders set the angle of incidence, the prism angle from 30 to 75 degrees and the refractive index from 1.30 to 1.80, and a button jumps to the angle of minimum deviation. Readouts give r1, r2, e, the deviation, the minimum deviation and the incidence at which it occurs.

Fig. 9.21Change the angle of incidence, the prism angle and the refractive index. The dot on the graph shows the current deviation; the dashed line marks the minimum.

At the bottom of the curve only one i gives the deviation, so the path must be its own reverse: e=i and r1=r2. The deviation there is the angle of minimum deviation Dm, and for a prism with equal sides the ray inside runs parallel to the base. Putting r1=r2=r and δ=Dm, i=e into (9.34) and (9.35):

r=A2,i=A+Dm2
(9.37)
refractive index from minimum deviation
n21=n2n1=sin⁡[(A+Dm)/2]sin⁡(A/2)
(9.38)

Both A and Dm can be measured precisely, for instance with a spectrometer, so (9.38) is the standard laboratory way to find the refractive index of a glass, and of each colour separately.

For a thin prism, with A only a few degrees, Dm is small too, and the sines can be replaced by their angles: n21≈(A+Dm)/2A/2, which gives

Dm≈(n21−1)A

A 4∘ prism of glass with n=1.5 deviates light by only about 2∘. Spectacles for people whose eyes do not point together sometimes include such weak prisms.

Example

Refractive index from minimum deviation

Light passes through a glass prism of angle 60∘. The angle of minimum deviation is measured as 46∘. Find the refractive index of the glass, the angle of incidence at minimum deviation, and the deviation produced by a 5∘ prism of the same glass.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A ray passes through a 60∘ prism with i=50∘ and e=40∘. What is the deviation, in degrees?
°
Show answer

Answer 30 °

δ=i+e−A=50∘+40∘−60∘=30∘. The same deviation would occur for i=40∘ (reversed path).

9.7 Optical instruments

Mirrors, lenses and prisms are the parts; optical instruments put them together. A periscope uses two mirrors or totally reflecting prisms to see over a wall; a kaleidoscope uses repeated reflections between mirrors to make patterns; binoculars combine lenses and prisms. The most important optical instrument of all is the human eye, which you studied in Class 10. Here we look at the two instruments that opened up the very small and the very far: the microscope and the telescope.

9.7.1 The microscope

How big a thing looks depends on the angle it subtends at your eye, not on its actual size: a coin at arm's length can hide the Moon. To see a small object better you bring it closer, but the eye cannot focus on anything nearer than the near point, about D=25 cm for a normal adult. So the largest angle a small object of height h can subtend, seen comfortably with the naked eye, is about h/D.

A simple microscope, or magnifying glass, is a converging lens of short focal length held close to the eye. The object is placed at or just inside the focal point. Then the lens forms an erect, enlarged, virtual image far enough away for the eye to focus on (Fig. 9.22).

2/3Object at F: the rays leave parallel at θi≈h/f, and a relaxed eye sees the image at infinity.

Fig. 9.22A magnifying glass lets the object come closer than 25 cm while its image stays at a comfortable distance.NCERT Fig. 9.23 · p. 241

An eye on the right looks along the axis. Step 1: with no lens, a small object stands 25 centimetres from the eye, at the near point; the line from its top to the eye makes a small angle theta zero with the axis. Step 2: a convex lens is placed just in front of the eye and the object is moved to its focus F; two rays from the object top, one parallel to the axis and one through the centre of the lens, leave the lens parallel to each other at a larger angle theta i, so the image is at infinity and a relaxed eye can view it. Step 3: the object is moved slightly inside F; the rays leaving the lens now diverge, and their dashed backward extensions meet at a large, upright, virtual image 25 centimetres from the eye, six times taller than the object.

DerivationMagnifying power of a simple microscope

Image at the near point. Multiply the lens formula 1/v−1/u=1/f by v: 1−v/u=v/f, so m=v/u=1−v/f. The virtual image is on the object side at the near point, v=−D:

image at the near point
m=1+Df
(9.39)

With D=25 cm, a lens of focal length 5 cm gives m=6. Because the image sits exactly at D, the same number compares its angular size with that of the object viewed at the near point.

Image at infinity. Viewing an image at the near point strains the eye, so the more relaxed arrangement puts the object at the focal point. Now the useful measure is the magnifying power, or angular magnification. Without the lens, the best angle is

θ0≈tan⁡θ0=hD
(9.40)

With the eye close to the lens and the object at the focus, rays from the top of the object leave parallel, inclined to the axis at

θi≈hf
(9.41)

So the angular magnification is

image at infinity
m=θiθ0=Df
(9.42)

This is one less than (9.39), a small price for relaxed viewing; from now on we take the final image at infinity. Notice what the lens really does. With the eye at the lens, the virtual image subtends the same angle as the object itself. The gain is that the object can be brought to a few centimetres from the eye, where it subtends a large angle, while its image stays at a distance the eye can focus on.

A simple microscope is limited to about 9 or 10 times. Shorter focal lengths need small, strongly curved lenses whose images are blurred and coloured at the edges. For more, we compound the magnification of two lenses.

A compound microscope splits the work between two lenses (Fig. 9.23). The objective, very close to the specimen, has a very short focal length; with the specimen just outside its focus, it throws a real, upside-down and already enlarged picture partway up the tube. The eyepiece then treats that picture the way a magnifying glass treats a postage stamp. The picture is made to fall at the eyepiece's focus (or slightly inside it), and the eye looks at it through the eyepiece, seeing a final image at infinity (or at the near point). Since nothing turns it back over, the specimen appears inverted, which hardly matters for a slide of cells.

2/3The first image sits at the eyepiece's focus; the eyepiece sends the rays out parallel, so the final image is at infinity.

Fig. 9.23The objective magnifies linearly; the eyepiece then magnifies angularly.NCERT Fig. 9.24 · p. 243

A small objective lens on the left with a tiny object just beyond its focus, and a larger eyepiece lens to the right. Step 1: two rays from the object top, one parallel to the axis and one through the centre of the objective, meet to form a large, inverted, real first image between the lenses. Step 2: from the top of this first image, which sits at the focus of the eyepiece, a ray parallel to the axis and a ray through the eyepiece centre leave the eyepiece parallel to each other, so the final image is at infinity. Step 3: the lengths are marked: f o, the focal length of the objective; L, the tube length from the second focus of the objective to the first focus of the eyepiece; and f e, the focal length of the eyepiece.

Magnification of the objective. Follow the ray that leaves the object's tip parallel to the axis. After the objective it passes through the second focal point and goes on to the tip of the first image. It makes one angle ϕ with the axis throughout, so the right triangle it forms before the focal point (height h, base fo) and the one after it (height h′, base L) are similar: tan⁡ϕ=h/fo=h′/L. Therefore

objective
mo=h′h=Lfo
(9.43)

Here L, the tube length, is the distance from the second focal point of the objective to the first focal point of the eyepiece. The eyepiece is a simple microscope, so its angular magnification is

final image at the near point
me=1+Dfe
(9.44a)
final image at infinity
me=Dfe
(9.44b)

By (9.33) the total magnification is the product. With the final image at infinity:

compound microscope
m=mome=Lfo⋅Dfe
(9.45)

Both focal lengths must be short for a large magnification. Lenses with focal lengths much below 1 cm are hard to make well, and a long tube makes the instrument unwieldy. Modern objectives and eyepieces each contain several lenses that cancel each other's defects, and good illumination of the specimen matters as much as magnification.

ExampleNCERT p. 244

A school microscope

A compound microscope has an objective of focal length 0.80 cm, an eyepiece of focal length 2.5 cm and a tube length of 16 cm. Find its magnifying power with the final image at infinity, and with the final image at the near point.

Try it yourself first. Then reveal the solution one step at a time.

9.7.2 Telescope

A telescope magnifies the angle between the directions of distant objects. In a refracting telescope the objective is a lens of long focal length and wide aperture. Light from a distant object arrives as parallel rays inclined at a small angle α to the axis, and the objective brings them to a real, inverted image in its focal plane. The eyepiece, placed one focal length beyond that image, sends the rays out parallel again, now inclined at a larger angle β (Fig. 9.24).

3/3The eyepiece sends the light out parallel at the larger angle β; the tube is fo+fe long.

Fig. 9.24The telescope turns a small angle α into a larger angle β, with β/α = fo/fe.NCERT Fig. 9.25 · p. 245

A large objective lens on the left and a small eyepiece lens on the right, on the same axis. Step 1: three parallel rays from a distant object arrive at a small angle alpha to the axis. Step 2: the objective brings them together at a point below the axis in its focal plane, where a small inverted real image forms. Step 3: the rays continue through the eyepiece, which is one eyepiece focal length beyond the image, and leave parallel to each other at a larger angle beta; the distances f o from the objective to the image and f e from the image to the eyepiece are marked.

If the first image has height h, the objective sees it at angle α≈h/fo, and the eyepiece sends rays from its tip out at β≈h/fe. The magnifying power is

telescope
m=βα≈h/feh/fo=fofe
(9.46)

In this normal adjustment the tube length is fo+fe. A telescope with fo=90 cm and fe=3.0 cm magnifies 30 times: two stars 0.2∘ apart in the sky appear 6∘ apart. An astronomical telescope shows things upside down, which does not matter for stars; a terrestrial telescope adds a pair of erecting lenses, and binoculars use totally reflecting prisms to turn the image upright and fold the long light path into a short case.

Example

A small refracting telescope

A refracting telescope has an objective of focal length 90 cm and an eyepiece of focal length 3.0 cm. Find its magnifying power and tube length in normal adjustment. Two moons of Jupiter are 0.10∘ apart as seen from the Earth. How far apart do they appear through the telescope?

Try it yourself first. Then reveal the solution one step at a time.

For astronomers, magnification is the cheap part: a different eyepiece changes it. Two things cannot be bought that way. A telescope collects starlight over the whole of its objective, so doubling the diameter gathers four times the light and reveals much fainter objects: its light-gathering power rises. A wider objective also shows two stars that lie almost in the same direction as two points instead of one blur: its resolving power improves. So the race has always been for bigger objectives. Lenses lose that race. A big glass disc is heavy, can be held only at its rim (where it sags), and must be flawless and free of colour defects through its whole thickness. The largest lens telescope ever built, finished in 1897 at Yerkes Observatory in the USA, has an objective about 1 m (40 inches) across.

Nearly every large research telescope is therefore a reflecting telescope, with a concave mirror as its objective; the comparison below sums up why. A mirror has one awkward feature: it sends the light back the way it came, so its focus lies in front of it, in the path of the incoming starlight. The early giants simply put the astronomer there. On the 5.08 m (200-inch) telescope at Mount Palomar in the USA, an observer rode in a small cage at the prime focus, shading a little of the mirror. The Cassegrain design is neater. A small convex secondary mirror catches the converging light before it reaches the focus and sends it back down the tube, through a hole in the middle of the primary, to an eyepiece or camera behind it (Fig. 9.25). Because the secondary also stretches the effective focal length, a long focal length fits into a short tube.

Fig. 9.25A convex secondary folds the light back through a hole in the primary: a long focal length in a short tube.NCERT Fig. 9.26 · p. 246

A short tube open on the left. Parallel rays enter and travel to a large concave primary mirror on the right, which has a hole at its centre. The mirror reflects them towards a focus, but before they get there a small convex secondary mirror near the open end reflects them back towards the primary. The rays pass through the central hole and meet at the image behind the primary, where an eyepiece is placed.

Compare

Refracting and reflecting telescopes

Refracting (lens objective)

  • light passes through glass, which must be flawless throughout
  • colour fringes (chromatic aberration)
  • heavy lens held only at its rim
  • practical limit about 1 m

Reflecting (mirror objective)

  • only the front surface must be perfect
  • no chromatic aberration
  • mirror supported across its whole back
  • mirrors of 8–10 m in use
Quick check
1 of 4
A telescope has an objective of focal length 1.2 m and an eyepiece of focal length 4.0 cm. What is its magnifying power in normal adjustment?
times
Show answer

Answer 30 times

m=fo/fe=120 cm/4.0 cm=30. Put both focal lengths in the same unit first.
Recap

Ray Optics and Optical Instruments: key points and summary

  1. Light is the visible band of electromagnetic waves, λ≈400–750 nm; because λ is tiny compared with everyday objects, it can be treated as travelling along straight rays (Section 9.1).
  2. The laws of reflection hold at every point of a curved mirror, with the normal along the radius. Cartesian convention: distances from the pole or optical centre, positive along the incident light; heights up positive (Section 9.2.1).
  3. For paraxial rays a spherical mirror has f=R/2; concave mirrors have f<0, convex mirrors f>0 (Section 9.2.2).
  4. Mirror equation 1/v+1/u=1/f and magnification m=h′/h=−v/u hold for every case when the signs are used (Section 9.2.3).
  5. Snell's law: sin⁡i/sin⁡r=n21=n2/n1, or n1sin⁡i=n2sin⁡r. A slab shifts a ray sideways without deviating it; for near-normal viewing, apparent depth = real depth / n (Section 9.3).
  6. Going from a denser to a rarer medium, light is totally reflected when i exceeds the critical angle, sin⁡ic=n2/n1 (Section 9.4).
  7. Totally reflecting 45∘ prisms turn or invert beams without loss; optical fibres guide light along a core of higher index than the cladding (Section 9.4.1).
  8. A spherical refracting surface obeys n2/v−n1/u=(n2−n1)/R (Section 9.5.1).
  9. Thin lens: 1/v−1/u=1/f, lens maker's formula 1/f=(n21−1)(1/R1−1/R2), magnification m=v/u (Section 9.5.2).
  10. Power P=1/f in dioptres; for thin lenses in contact powers add and magnifications multiply (Section 9.5.3, Section 9.5.4).
  11. A prism gives r1+r2=A and δ=i+e−A; at minimum deviation i=e and n21=sin⁡[(A+Dm)/2]/sin⁡(A/2); a thin prism deviates by (n−1)A (Section 9.6).
  12. A simple microscope magnifies 1+D/f (image at D=25 cm) or D/f (image at infinity); a compound microscope magnifies about (L/fo)(D/fe) (Section 9.7.1).
  13. A telescope has angular magnification fo/fe and length fo+fe in normal adjustment; large telescopes use mirrors to avoid chromatic aberration and to support big objectives (Section 9.7.2).
Flash cards · 12

Formula revision

1 / 12
Practice · 36 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises 9.1

    In a physics lab, a concave mirror with a radius of curvature of 40 cm is used to project the image of a glowing LED, 3.0 cm tall, onto a white card. The LED is 30 cm from the mirror.

    1. Where must the card be held for a sharp picture?

      cm
      Show solution

      f=−20 cm, u=−30 cm: 1/v=1/f−1/u=−1/20+1/30=−1/60, so v=−60 cm: 60 cm in front of the mirror.

    2. How tall is the picture, and which way up?

      Show solution

      m=−v/u=−(−60)/(−30)=−2: real, inverted and 6.0 cm tall.

    3. The LED is now pushed towards the mirror, a little at a time. How must the card be moved to keep the picture sharp, and what happens once the LED is nearer than 20 cm?

      Show solution

      As the LED approaches F (20 cm) the image distance grows without limit, so the card must be moved further and further back. Inside F the image becomes virtual, behind the mirror, and no card position catches it.

  2. NumericalEasyNCERT Exercises 9.2

    A child holds a toy car 2.0 cm tall 30 cm in front of the back of a polished steel ladle, which acts as a convex mirror of focal length 20 cm.

    1. Where is the image of the toy, and how tall is it?

      cm
      Show solution

      f=+20 cm, u=−30 cm: 1/v=1/20+1/30=1/12, so v=+12 cm, behind the ladle.

      m=−v/u=−12/(−30)=+0.40: virtual, erect, 0.80 cm tall.

    2. The child slowly moves the toy further away. What happens to the image?

      Show solution

      As u→−∞, 1/v→1/f: the image slides towards F, 20 cm behind the mirror, and keeps shrinking, but it stays virtual and erect.

  3. NumericalMediumNCERT Exercises 9.3

    A coin lies at the bottom of a beaker under 24.0 cm of water (n=1.33). A travelling microscope looks down at it from above.

    1. At what depth below the water surface does the microscope see the coin?

      cm
      Show solution

      24.0/1.33=18.0 cm below the surface.

    2. The water is replaced, to the same depth, by an unknown liquid. To refocus on the coin the microscope has to be raised by 3.0 cm. Find the refractive index of the liquid.

      (no unit)
      Show solution

      The coin now appears 3.0 cm higher, at an apparent depth of 15.0 cm. So n=24.0/15.0=1.60.

  4. NumericalMediumNCERT Exercises 9.4

    A ray from air entering a glass block at 50∘ to the normal refracts to 29.0∘. A ray from air entering water at 50∘ refracts to 35.2∘. An aquarium has a glass floor; a ray travelling in its water meets the glass at 40∘ to the normal. At what angle does it travel in the glass?

    °
    Show solution

    ng=sin⁡50∘/sin⁡29.0∘=0.766/0.485=1.58; nw=sin⁡50∘/sin⁡35.2∘=0.766/0.576=1.33.

    At the water–glass boundary: sin⁡r=(nw/ng)sin⁡40∘=(1.33/1.58)(0.643)=0.541, so r=32.7∘, bent slightly towards the normal.

  5. NumericalMediumNCERT Exercises 9.5

    At night, a drone hovering over a swimming pool photographs a lamp lying on the pool floor. The water surface shows a sharply bounded bright circle 4.6 m across, with darkness outside it (n=1.33).

    1. Why does the bright patch have a sharp edge?

      Show solution

      Light from the lamp reaching the surface outside the circle meets it at more than the critical angle (48.8∘) and is totally reflected back into the pool. Inside the circle some light refracts out. The switch from 'some escapes' to 'none escapes' happens abruptly at ic.

    2. How deep is the pool?

      m
      Show solution

      ic=sin⁡−1(1/1.33)=48.8∘, tan⁡ic=1.14. Radius =2.3 m, so depth =2.3/1.14=2.0 m.

  6. NumericalMediumNCERT Exercises 9.6

    A student measures the angle of minimum deviation of a 60∘ glass prism in air as 44∘. She then immerses the prism in a tank of water (n=1.33).

    1. What refractive index does her measurement give?

      (no unit)
      Show solution

      n=sin⁡[(60∘+44∘)/2]/sin⁡30∘=sin⁡52∘/0.500=1.58.

    2. What minimum deviation should she expect in water?

      °
      Show solution

      Relative index =1.576/1.33=1.185. Then sin⁡[(60∘+Dm)/2]=1.185×0.500=0.5925, so (60∘+Dm)/2=36.3∘ and Dm=12.7∘.

    3. What would the prism do to light in a clear oil of index 1.58?

      Show solution

      The relative index is 1.576/1.58≈1.00, so light would hardly bend at all: the prism would almost vanish.

  7. NumericalEasyNCERT Exercises 9.7

    A lens maker needs equiconvex lenses of focal length 15 cm and has two glasses, of refractive index 1.50 and 1.60. What radius of curvature must the faces have for each glass, and which glass gives the flatter (thinner) lens?

    cm
    Show solution

    R=2(n−1)f. For n=1.50: R=2×0.50×15=15 cm. For n=1.60: R=2×0.60×15=18 cm.

    The higher-index glass needs less curved faces (18 cm), so it gives the thinner lens.

  8. NumericalMediumNCERT Exercises 9.8

    A projector lens sends a beam converging towards a point P on a screen. A second lens is slipped into the beam 15 cm before P.

    1. Where does the beam now converge if the new lens is convex with f=30 cm?

      cm
      Show solution

      1/v=1/f+1/u=1/30+1/15=1/10, so v=+10 cm: 10 cm beyond the lens, nearer than P.

    2. And if it is concave with f=20 cm?

      cm
      Show solution

      1/v=−1/20+1/15=1/60, so v=+60 cm: still converging, but now to a point 60 cm beyond the lens.

    3. What lens, placed at the same spot, would make the beam come out parallel?

      Show solution

      Parallel light means v→∞: 1/f=−1/u=−1/15, so a diverging lens of focal length 15 cm (f=−15 cm).

  9. NumericalEasyNCERT Exercises 9.9

    A diverging lens with f=−15 cm is used to view a 2.0 cm high object placed 30 cm away from it. Describe the image fully. Can moving the object ever make this lens's image real, or larger than the object?

    cm
    Show solution

    1/v=1/f+1/u=−1/15−1/30=−1/10, so v=−10 cm: on the same side as the object.

    m=v/u=(−10)/(−30)=+1/3: virtual, erect, 0.67 cm high.

    For any real object both terms in 1/f+1/u are negative, so v is always negative (virtual) and ∣v∣<∣u∣, giving 0<m<1. Moving the object away only pushes the image towards F and shrinks it.

  10. NumericalEasyNCERT Exercises 9.10

    Two thin lenses touch: a converging one with f=+25 cm and a diverging one with f=−40 cm. Find the focal length of the pair. Does it converge or diverge light?

    cm
    Show solution

    1/f=1/25−1/40=(8−5)/200=3/200, so f=+66.7 cm.

    Positive: the pair is a weak converging lens, because the converging lens is the stronger of the two.

  11. NumericalHardNCERT Exercises 9.11

    A microscope's objective (f=1.5 cm) and eyepiece (f=5.0 cm) are fixed 16 cm apart, and it is focused by moving the slide. A student first focuses for relaxed viewing (final image at infinity), then refocuses so that the final image is at her near point, 25 cm.

    1. How far from the objective is the slide in each setting?

      cm
      Show solution

      Infinity: the first image must be at the eyepiece focus, 16−5.0=11.0 cm behind the objective. 1/uo=1/11.0−1/1.5 gives uo=−1.737 cm.

      Near point: 1/ue=−1/25−1/5.0 gives ue=−4.17 cm, so the first image is at 16−4.17=11.83 cm. 1/uo=1/11.83−1/1.5 gives uo=−1.718 cm.

      The slide moves only about 0.19 mm closer to the objective between the two settings.

    2. Find the magnifying power in each setting.

      times
      Show solution

      Infinity: ∣mo∣=11.0/1.737=6.33, me=25/5.0=5.0, so m≈32.

      Near point: ∣mo∣=11.83/1.718=6.89, me=1+25/5.0=6.0, so m≈41.

  12. NumericalHardNCERT Exercises 9.12

    In a student's microscope the objective has f=6.0 mm and the eyepiece f=3.0 cm. A cell sample 7.0 mm below the objective is seen sharply, with the final image at the near point (25 cm).

    1. How far apart are the lenses, and what is the magnifying power?

      times
      Show solution

      Objective: 1/vo=1/0.60−1/0.70=0.238 cm−1, so vo=4.2 cm and ∣mo∣=4.2/0.70=6.0.

      Eyepiece: 1/ue=−1/25−1/3.0, so ue=−2.68 cm. Separation =4.2+2.68=6.9 cm.

      Magnifying power =6.0×(1+25/3.0)=6.0×9.33=56.

    2. To view with a relaxed eye instead, should the eyepiece be moved in or out, and by how much? What is the magnifying power then?

      times
      Show solution

      For a final image at infinity the first image must sit at the eyepiece focus, so the separation becomes 4.2+3.0=7.2 cm: move the eyepiece out by about 0.32 cm.

      Magnifying power =6.0×25/3.0=50.

  13. NumericalEasyNCERT Exercises 9.13

    A birdwatcher's refracting telescope is 124 cm long in normal adjustment and magnifies 30 times. Find the focal lengths of its objective and eyepiece.

    cm
    Show solution

    fo=30fe, so 31fe=124 cm: fe=4.0 cm and fo=120 cm.

  14. NumericalMediumNCERT Exercises 9.14

    A solar observatory uses a lens objective of focal length 12 m.

    1. Paired with a 2.0 cm eyepiece, what angular magnification would it give?

      times
      Show solution

      m=fo/fe=1200 cm/2.0 cm=600.

    2. Instead of an eyepiece, a card is placed at the focus of the objective to catch the Sun's image (no one looks through it). The Sun's diameter is 1.39×109 m and its distance 1.50×1011 m. How wide is the Sun's image?

      cm
      Show solution

      Angle subtended: α=1.39×109/1.50×1011=9.27×10−3 rad. Image width =foα=12×9.27×10−3=0.111 m≈11 cm.

  15. ProofMediumNCERT Exercises 9.15

    Use the mirror equation 1/v+1/u=1/f, with signs, to show each of the following without drawing rays.

    1. A concave mirror forms a real image between F and C of an object placed beyond C.

      Show solution

      Here f<0 and u<2f (more negative). Then 1/u lies between 1/(2f) and 0, so 1/v=1/f−1/u lies between 1/f and 1/(2f), both negative. Hence v is negative and lies between f and 2f: a real image between F and C.

    2. A convex mirror always forms a virtual image of a real object.

      Show solution

      f>0 and u<0, so 1/v=1/f−1/u=1/f+1/∣u∣>0: v is always positive, behind the mirror. The image is virtual.

    3. That virtual image is always smaller than the object and lies between the pole and the focus.

      Show solution

      Since 1/v=1/f+1/∣u∣ is larger than both 1/f and 1/∣u∣, the image distance v is smaller than both f and ∣u∣. So 0<v<f: the image lies between P and F. And m=−v/u=v/∣u∣ lies between 0 and 1: erect and diminished.

    4. A concave mirror forms a virtual, enlarged image of an object placed between the pole and the focus.

      Show solution

      f<0 and f<u<0, so ∣u∣<∣f∣ and −1/u=1/∣u∣>1/∣f∣. Then 1/v=−1/∣f∣+1/∣u∣>0, so v>0: virtual. Also 1/v<1/∣u∣, so v>∣u∣ and m=v/∣u∣>1: enlarged and erect.

  16. NumericalMediumNCERT Exercises 9.16

    A letter lies under a glass paperweight that is a flat slab 12 cm thick (n=1.60), and you look straight down at it.

    1. By how much does the letter appear to be raised?

      cm
      Show solution

      The 12 cm of glass looks only 12/1.60=7.5 cm thick, so the letter appears raised by 12−7.5=4.5 cm.

    2. The same slab is now held 10 cm above the letter instead of resting on it. How far does the letter appear raised now?

      cm
      Show solution

      Still 4.5 cm. Only the glass changes the apparent distance, by t(1−1/n); the air gaps above and below it look as long as they are. So the shift does not depend on where the slab is, for near-normal viewing.

  17. NumericalHardNCERT Exercises 9.17

    A fibre in a medical endoscope has a glass core of index 1.62 coated with a glass of index 1.50.

    1. What is the widest angle to the fibre's axis at which light from air can enter the flat end and still be trapped in the core?

      °
      Show solution

      sin⁡ic=1.50/1.62=0.926, ic=67.8∘. Inside, the ray may make at most 90∘−67.8∘=22.2∘ with the axis.

      At the end face: sin⁡θ=1.62sin⁡22.2∘=0.612, so θ=37.7∘. Light within about 38∘ of the axis is guided.

    2. How would this change if the coating were removed and the core were surrounded by air?

      Show solution

      Now sin⁡ic=1/1.62, ic=38.1∘, so the angle with the axis inside may reach 51.9∘. The matching outside angle would need sin⁡θ=1.62sin⁡51.9∘=1.27>1, so light entering the end at any angle up to 90∘ is trapped. (A bare fibre is easily scratched, and anything touching it lets light leak out, which is why fibres are coated.)

  18. NumericalMediumNCERT Exercises 9.18

    A slide projector must throw an image of a slide onto a screen 2.4 m away from the slide, using a single convex lens. What is the largest focal length the lens can have?

    m
    Show solution

    For a real image with object and screen a distance D apart, u and v must satisfy ∣u∣+v=D and 1/v+1/∣u∣=1/f. Eliminating v gives a quadratic in ∣u∣ that has real solutions only if D≥4f.

    So fmax⁡=D/4=2.4/4=0.60 m; then the slide and screen are each 1.2 m from the lens (u=−2f, v=2f).

  19. NumericalMediumNCERT Exercises 9.19

    A lamp and a screen stand 80 cm apart, and a convex lens of focal length 15 cm is slid along the bench between them.

    1. At what distances from the lamp does the lens give a sharp image on the screen?

      cm
      Show solution

      ∣u∣(80−∣u∣)=15×80=1200, so ∣u∣2−80∣u∣+1200=0, giving ∣u∣=20 cm or 60 cm. The two positions are 40 cm apart.

    2. How do the two images compare?

      Show solution

      At ∣u∣=20, v=60: m=−3. At ∣u∣=60, v=20: m=−1/3. Both are real and inverted; one is three times the lamp's size, the other a third. Object and image distances simply swap.

    3. Show that, in general, if two sharp positions a distance d apart exist for a lamp–screen separation D, then f=(D2−d2)/4D, and check it here.

      Show solution

      The two positions swap ∣u∣ and v, so ∣u∣+v=D and v−∣u∣=d, giving ∣u∣=(D−d)/2, v=(D+d)/2. Then 1/f=1/v+1/∣u∣=4D/(D2−d2). Here (6400−1600)/320=15 cm.

  20. NumericalHardNCERT Exercises 9.20

    Take the two lenses of Question 10 (converging, f=25 cm; diverging, f=40 cm) and mount them 10 cm apart on a common axis.

    1. A distant street light is viewed through the pair. Where is its image if the converging lens faces the light? And if the diverging lens faces it? Can the pair be described by a single focal length?

      Show solution

      Converging lens first: it would focus 25 cm behind itself, 15 cm beyond the diverging lens, so u2=+15 cm: 1/v=−1/40+1/15=1/24, v=24 cm behind the diverging lens.

      Diverging lens first: the light spreads from a point 40 cm in front of it, 50 cm in front of the converging lens: 1/v=1/25−1/50=1/50, so v=50 cm behind the converging lens.

      The focus lies at a different distance from the last lens depending on which way the light goes, so no single thin lens in a fixed place can replace the pair. (A combined focal length, 1/F=1/f1+1/f2−d/f1f2, which gives 40 cm, exists only if distances are measured from special reference planes.)

    2. A 2.0 cm object stands 75 cm in front of the converging lens. Find the position, nature and size of the final image.

      cm
      Show solution

      Converging lens: 1/v1=1/25−1/75=2/75, v1=37.5 cm, m1=37.5/(−75)=−0.50.

      The diverging lens is 10 cm further on, so u2=37.5−10=+27.5 cm (virtual object): 1/v2=−1/40+1/27.5, v2=88 cm, m2=88/27.5=3.2.

      m=(−0.50)(3.2)=−1.6: real, inverted, 3.2 cm high, 88 cm behind the diverging lens.

  21. NumericalMediumNCERT Exercises 9.21

    Light is to be sent into one face of a 60∘ prism made of glass of index 1.50 so that some of it escapes through the second face. What is the smallest angle of incidence on the first face for which any light gets out?

    °
    Show solution

    sin⁡ic=1/1.50, ic=41.8∘. The limiting ray has r2=41.8∘, so r1=60∘−41.8∘=18.2∘.

    sin⁡i=1.50sin⁡18.2∘=0.468, so i=27.9∘. At smaller angles of incidence the light is totally reflected inside at the second face; at this angle it just grazes out.

  22. NumericalMediumNCERT Exercises 9.22

    A watch repairer examines a gear whose teeth are 2.0 mm apart through a loupe (a magnifying lens) of focal length 12 cm held at the eye, with the gear 10 cm from the lens.

    1. Where is the image, and how far apart do the teeth appear in it?

      mm
      Show solution

      1/v=1/12+1/(−10)=−1/60, v=−60 cm; m=v/u=(−60)/(−10)=6.0. The image is 60 cm away, with teeth 12 mm apart.

    2. What is the magnifying power?

      times
      Show solution

      Seen from the lens, the image spacing subtends 12 mm/600 mm=0.020 rad. With the naked eye at 25 cm the gear spacing subtends 2.0/250=0.0080 rad. Magnifying power =0.020/0.0080=2.5.

    3. Why are the answers different?

      Show solution

      The image is six times larger, but it is also far away (60 cm), so it does not look six times larger. Apparent size is set by the angle at the eye, which is what the magnifying power compares. The two agree only when the image is at the near point (Question 23).

  23. NumericalMediumNCERT Exercises 9.23

    The repairer of Question 22 wants the loupe to make the gear look as large as possible while the image stays sharp.

    1. How far from the lens should the gear be?

      cm
      Show solution

      Image at v=−25 cm: 1/u=1/v−1/f=−1/25−1/12, so u=−8.1 cm.

    2. What is the linear magnification then?

      times
      Show solution

      m=v/u=25/8.11=3.08.

    3. Does it now equal the magnifying power? Why?

      Show solution

      Yes. The image is at 25 cm, exactly where the naked-eye comparison is made, so comparing angles is the same as comparing sizes: both equal 1+D/f=3.08.

  24. NumericalMediumNCERT Exercises 9.24

    With the same loupe (f=12 cm), where should the gear be placed so that its teeth appear 6.0 mm apart in the image? Could the repairer see them sharply with the loupe right at the eye?

    cm
    Show solution

    m=6.0/2.0=3, so v=3u. Then 1/(3u)−1/u=1/12 gives −2/(3u)=1/12, so u=−8.0 cm and v=−24 cm.

    The image would be 24 cm away, just inside the 25 cm near point, so most people would see it slightly blurred or find it tiring.

  25. Short answerMediumNCERT Exercises 9.25

    Answer briefly.

    1. Seen from right behind a hand lens, the enlarged image and the object itself make equal angles at the eye. So what does the lens actually gain you?

      Show solution

      Without the lens the object cannot be brought closer than 25 cm and still be seen sharply. The lens lets you hold it a few centimetres away, where it subtends a much larger angle, while its image sits at a distance the eye can focus on.

    2. You hold a hand lens fixed above a stamp and pull your head back from it. Does the magnifying power stay the same?

      Show solution

      Yes, it falls. The angle subtended by the image at the eye decreases as the eye moves back, and less of the field is visible.

    3. Since D/f keeps growing as f shrinks, why do shops not sell 100× hand lenses?

      Show solution

      Short focal lengths need small, strongly curved lenses. Their spherical and chromatic aberrations grow so fast that the image becomes blurred and coloured, so simple magnifiers stop at about 10×.

    4. A microscope maker picks short-focus lenses for both ends of the tube. Why both, and not just one?

      Show solution

      m≈(L/fo)(D/fe): the objective's linear magnification and the eyepiece's angular magnification both grow as their focal lengths shrink.

    5. Through a microscope, the view is best with the eye held slightly back from the eyepiece rather than pressed against it. Why, and how far back?

      Show solution

      All the light passing through the objective converges, after the eyepiece, through a small circle just behind it, the image of the objective formed by the eyepiece (the eye ring). With the pupil there the eye collects all the light and sees the full field. For typical instruments it is a little more than fe behind the eyepiece, about a centimetre or two.

  26. NumericalHardNCERT Exercises 9.26

    A kit contains an objective lens (f=1.0 cm) and an eyepiece (f=5.0 cm).

    1. Design a microscope from them that magnifies 25 times with the final image at the near point: where must the object be, and how far apart must the lenses be?

      cm
      Show solution

      me=1+25/5.0=6, so ∣mo∣=25/6=4.17.

      With vo=4.17∣uo∣ in 1/vo+1/∣uo∣=1/fo: (1+4.17)/(4.17∣uo∣)=1, so ∣uo∣=1.24 cm and vo=5.17 cm.

      Eyepiece: 1/ue=−1/25−1/5.0, so ∣ue∣=4.17 cm. Lens separation =5.17+4.17=9.33 cm; object 1.24 cm from the objective.

    2. Keeping the lenses 9.33 cm apart, the user refocuses for a relaxed eye (final image at infinity). What is the magnifying power now?

      times
      Show solution

      The first image must now be at the eyepiece focus, 9.33−5.0=4.33 cm from the objective: 1/∣uo∣=1−1/4.33, ∣uo∣=1.30 cm, ∣mo∣=4.33/1.30=3.33.

      m=3.33×25/5.0=16.7.

  27. NumericalMediumNCERT Exercises 9.27

    A hobby telescope pairs an objective of focal length 150 cm with an eyepiece of focal length 6.0 cm. It is first used with a relaxed eye and then refocused so that the final image of a distant hill is at 25 cm.

    1. What is the magnifying power with a relaxed eye?

      times
      Show solution

      m=150/6.0=25.

    2. What is it with the final image at 25 cm, and by what factor has it changed?

      times
      Show solution

      m=(fo/fe)(1+fe/D)=25×(1+6.0/25)=31, larger by a factor 1.24.

  28. NumericalMediumNCERT Exercises 9.28

    The telescope of Question 27 is pointed at a water tank 60 m tall.

    1. How far apart are its lenses with a relaxed eye?

      cm
      Show solution

      fo+fe=150+6.0=156 cm.

    2. The objective forms an image of the tank 4.5 cm tall. How far away is the tank?

      km
      Show solution

      The tank subtends 4.5 cm/150 cm=0.030 rad, so its distance is 60 m/0.030=2.0×103 m, i.e. 2.0 km.

    3. With a relaxed eye, what angle does the tank appear to fill?

      °
      Show solution

      25×0.030 rad=0.75 rad, about 43∘ (the small-angle approximation is already stretched here).

  29. NumericalHardNCERT Exercises 9.29

    A small Cassegrain telescope has a concave primary mirror of radius of curvature 200 mm and a convex secondary mirror of radius 60 mm, placed 75 mm in front of the primary. Where is the final image of a star formed?

    mm
    Show solution

    Primary: f1=100 mm, so the reflected light converges towards a point 100 mm in front of it, 25 mm beyond the secondary.

    For the secondary, measured along the light now travelling towards it: u=+25 mm (virtual object behind it) and f2=+30 mm (convex).

    1/v=1/f2−1/u=1/30−1/25=−1/150, so v=−150 mm: a real image 150 mm in front of the secondary, back towards the primary.

    That is 150−75=75 mm behind the primary mirror, where the light emerges through its central hole. The secondary magnifies by 6, so the system acts like a 600 mm focal length packed into a short tube.

  30. NumericalMediumNCERT Exercises 9.30

    In a sensitive galvanometer, a small plane mirror on the coil reflects a light beam back onto a scale 1.2 m away; with no current the beam strikes the mirror normally and returns on itself. When a current flows, the light spot moves 8.4 cm along the scale. Through what angle has the coil turned?

    °
    Show solution

    The reflected beam has turned through tan⁡−1(0.084/1.2)=4.0∘. The mirror, and so the coil, turned through half of that: 2.0∘.

    The doubling is why a mirror and a long light lever make tiny rotations easy to read.

  31. NumericalHardNCERT Exercises 9.31

    An equiconvex glass lens (n=1.50) rests on a plane mirror. A pin is moved along the axis above it until the pin and its inverted image coincide; this happens 24.0 cm above the lens.

    1. What is the radius of curvature of the lens faces?

      cm
      Show solution

      f=24.0 cm and 1/f=0.50×2/R, so R=24.0 cm.

    2. A few drops of a liquid are put between the lens and the mirror, and the coincidence distance becomes 40.0 cm. Find the liquid's refractive index.

      (no unit)
      Show solution

      The combination has F=40.0 cm, so the liquid lens has 1/fl=1/40.0−1/24.0=−1/60, fl=−60 cm.

      For the plano-concave liquid lens 1/fl=−(nl−1)/R, so nl−1=24.0/60=0.40 and nl=1.40.

    3. Where would the pin coincide if the liquid were water (n=1.33)?

      cm
      Show solution

      1/fl=−0.33/24.0, fl=−72.7 cm; 1/F=1/24.0−1/72.7, so F=35.8 cm.

  32. Quick answerMedium

    A concave mirror forms an image three times the size of a real object, on a screen. If the object is 20 cm from the mirror, the focal length of the mirror is

    Show answer

    Answer (A) −15 cm

    u=−20 cm, v=−60 cm: 1/f=−1/60−1/20=−1/15. Taking m=+3 (virtual) would give −30 cm, but a virtual image cannot be caught on a screen.
  33. Quick answerHard

    A crown-glass lens (n=1.52) has a focal length of 20 cm in air. What is its power, in dioptres, when it is in water (n=1.33)?

    D
    Show answer

    Answer 1.37 D

    fw=20×0.52/(1.52/1.33−1)=72.8 cm, so P=1/0.728=1.37 D.
  34. Quick answerEasy

    Put the stages of image formation in a compound microscope in order, starting from the object.

    1. Light from the object enters the objective
    2. That image lies at or within the focus of the eyepiece
    3. The eyepiece forms a final, virtual, magnified image
    4. The objective forms a real, inverted, enlarged image
    Show answer
    Answer
    1. Light from the object enters the objective
    2. The objective forms a real, inverted, enlarged image
    3. That image lies at or within the focus of the eyepiece
    4. The eyepiece forms a final, virtual, magnified image
    The objective's real image is the object for the eyepiece, which acts as a magnifying glass.
  35. Quick answerEasy

    Match each device with the effect it relies on.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • Endoscope : Total internal reflection in fibres
    • Rear-view mirror of a car : Wide field of a convex mirror
    • Shaving mirror : Enlarged virtual image in a concave mirror
    • Spectrometer measuring n : Minimum deviation in a prism
    Each device uses one idea from this chapter.
  36. Quick answerEasy

    Light can be totally internally reflected only when it travels from the optically medium towards the optically medium.

    Show answer

    Answer (1) denser, (2) rarer

    Only when bending away from the normal can the refracted ray reach 90∘; that happens going from a denser to a rarer medium.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

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

Quick check: MCQs on Ray Optics and Optical Instruments

  1. 9.1 Introduction3 questions
  2. 9.2.3 The mirror equation4 questions
  3. 9.3 Refraction4 questions
  4. 9.4.1 Total internal reflection in nature and its technological applications4 questions
  5. 9.5.4 Combination of thin lenses in contact4 questions
  6. 9.6 Refraction through a prism4 questions
  7. 9.7.2 Telescope4 questions
Glossary · 15

Key terms (English–Hindi)

angle of minimum deviationन्यूनतम विचलन कोणnyuntam vichalan kon
The smallest deviation Dm a prism can produce, reached when the ray passes symmetrically (i=e).First seen in the chapter →
Cartesian sign conventionकार्तीय चिह्न परिपाटीkartiya chihn paripati
Distances are measured from the pole or optical centre, positive along the incident light; heights above the axis are positive.First seen in the chapter →
critical angleक्रांतिक कोणkrantik kon
The angle of incidence in the denser medium for which the refracted ray grazes the boundary: sin⁡ic=n2/n1.First seen in the chapter →
focal lengthफोकस दूरीphokas doori
The distance from the pole of a mirror or the optical centre of a lens to its principal focus, f.First seen in the chapter →
magnificationआवर्धनaavardhan
The ratio of image height to object height, m=h′/h; negative for an inverted image.First seen in the chapter →
magnifying powerआवर्धन क्षमताaavardhan kshamata
How many times larger the angle at the eye becomes with the instrument: the final image's angle divided by the angle of the object viewed directly (at the near point, for a microscope).First seen in the chapter →
optical fibreप्रकाशिक तंतुprakashik tantu
A thin glass or plastic thread with a core of higher refractive index than its cladding, guiding light by total internal reflection.First seen in the chapter →
paraxial rayउपाक्षीय किरणupakshiya kiran
A ray close to the principal axis and making a small angle with it.First seen in the chapter →
power of a lensलेंस की क्षमताlens ki kshamata
P=1/f with f in metres, measured in dioptres (D); positive for converging lenses.First seen in the chapter →
principal focusमुख्य फोकसmukhya phokas
The point where rays parallel to the principal axis meet, or seem to diverge from, after reflection or refraction.First seen in the chapter →
rayकिरणkiran
A line along which light energy travels; a bundle of rays is a beam.First seen in the chapter →
real imageवास्तविक प्रतिबिंबvastavik pratibimb
An image where rays from an object point actually meet; it can be caught on a screen.First seen in the chapter →
refractive indexअपवर्तनांकapvartanank
For a medium, n=c/v; for a pair, n21=n2/n1=sin⁡i/sin⁡r.First seen in the chapter →
total internal reflectionपूर्ण आंतरिक परावर्तनpoorn aantarik paravartan
Complete reflection of light inside a denser medium when it meets a rarer medium beyond the critical angle.First seen in the chapter →
virtual imageआभासी प्रतिबिंबaabhasi pratibimb
An image from which rays only appear to diverge; it cannot be caught on a screen.First seen in the chapter →

You reached the end