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PhysicsCh 10 Wave Optics

Wave Optics: Class 12 Physics · Chapter 10

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

Published · Updated

  • 95 min
  • 15 practice questions
  • 26 MCQs

What you will learn in Wave Optics

  • Explain what a wavefront is and use Huygens' construction to find later positions of spherical and plane wavefronts
  • Derive the laws of refraction and reflection from Huygens' principle and explain why speed and wavelength change at a boundary while frequency does not
  • Relate the critical angle to refractive indices and describe how prisms, lenses and mirrors reshape wavefronts
  • Distinguish coherent from incoherent sources and use path and phase difference to predict interference, including I = 4I0 cos²(φ/2)
  • Explain Young's double-slit experiment and calculate fringe positions and fringe width
  • Describe single-slit diffraction, locate its minima, and explain how it shapes a real double-slit pattern
  • Explain polarisation as a property of transverse waves, describe how polaroids work and apply Malus' law

Original companion text aligned to NCERT Physics Class 12, Chapter 10, pp. 255–273. Not affiliated with or endorsed by NCERT.

10.1 Introduction

Look at a soap bubble in sunlight, or at the sheen of spilt oil on a wet road, and you see bands of colour that no prism made. Hold two fingers almost touching in front of a tube light and squint through the gap: thin dark lines appear in it. Ray optics, which treats light as straight lines, explains neither. In this chapter we treat light as a wave, and that single idea accounts for reflection and refraction as well as for these new effects.

Is light a stream of particles or a wave? Both pictures explain why a ray reflects at equal angles and why it bends at a water surface, so for well over a century neither side could prove the other wrong. Fig. 10.1 shows the one place where they part company: how fast light travels inside water.

4/4Foucault's 1850 measurement: light is slower in water, as the wave model says.

Fig. 10.1Same bending, opposite predictions. Measurement settled it in favour of waves.NCERT 10.1 · p. 255

A ray of light passes from air (top) into water (bottom), bending towards the dashed normal; the angles i and r are marked. Step 1 shows only the bent ray. Step 2 adds the particle model on the right: two velocity arrows with dashed components, the one in water having the same sideways part but a longer downward part, so the model predicts that light is faster in water. Step 3 replaces this with short blue wavefronts drawn across the ray, spaced more closely in the water, and the label that the wave model predicts slower light. Step 4 keeps the wavefronts and adds a box with the measured speeds: about 3.00 times ten to the 8 metres per second in air and 2.25 times ten to the 8 in water, so the wave model wins.

In the particle picture, a particle of light reaching water is tugged towards the surface. The tug adds to the part of its velocity perpendicular to the surface and leaves the part along the surface alone, so the particle swings towards the normal and speeds up. In the wave picture, the end of a wavefront that enters the water first is slowed down, the rest of the front is still moving fast, and so the whole front pivots towards the normal. The bending is the same; the predictions about speed are opposite.

Timing light over a few metres of water was impossible until 1850, when Léon Foucault bounced a beam off a mirror spinning hundreds of times a second, once through air and once through a tube of water. The returning spot was displaced more for the water path, so light had taken longer there. Light is slower in water: about 2.25×108 m/s against 3.00×108 m/s in air. The wave model had passed a test that the particle model failed.

Timeline

Particles or waves?

  1. 1637

    Descartes

    Pictures light as particles and derives the law of refraction from that picture.

  2. 1678

    Huygens

    Proposes a wave theory and the construction with secondary wavelets (printed in 1690).

  3. 1704

    Newton's Opticks

    Newton's prestige makes the particle (corpuscular) model the accepted view for a century.

  4. 1801

    Young

    Two-slit interference: light from one source, split in two, gives bright and dark bands.

  5. 1818

    Fresnel

    A wave theory of diffraction that predicts the patterns at edges and slits in detail.

  6. 1850

    Foucault

    Light is slower in water than in air, as the wave model predicts.

  7. 1865

    Maxwell

    Light is an electromagnetic wave and needs no material medium.

  8. 1887

    Hertz

    Makes and detects electromagnetic waves at radio wavelengths.

  9. 1895

    J. C. Bose

    Works with millimetre-length electromagnetic waves in Calcutta.

By then the wave model had other evidence. From 1801 Thomas Young showed that light from one source, split into two beams that overlap again, gives bright and dark bands, the hallmark of waves (Section 10.5). Such experiments also measured the wavelength of light, and it turned out to be tiny: visible light spans roughly 400 to 700 nm, so about 150 wavelengths of green light fit across a human hair. Next to a mirror or a lens a few centimetres wide, a wave this short behaves almost exactly like a bundle of straight rays. Ray optics is what wave optics becomes when the wavelength is treated as vanishingly small, and a ray is then simply the direction in which the wave carries its energy.

One objection remained. Every wave known at the time, from ripples to sound, travels through some material, so how could light cross the empty space between the Sun and the Earth? Maxwell's equations answered it in the 1860s: light is an electromagnetic wave, a pattern of electric and magnetic fields that sustain each other and need no medium (Part I, Chapter 8). Hertz produced such waves at radio wavelengths in 1887, and J. C. Bose and G. Marconi soon put them to practical use.

Four questions organise the rest of the chapter. How does a wave model reproduce the laws of reflection and refraction (Section 10.2, Section 10.3)? What happens where two waves overlap, in general (Section 10.4) and in Young's experiment (Section 10.5)? What does a wave do when it squeezes through a narrow opening (Section 10.6)? And how can we tell that light waves are transverse (Section 10.7)?

Quick check
1 of 3
Light bends towards the normal as it passes from air into water. What did the particle model predict about its speed in water?
Show answer

Answer (A) Greater than in air

In the particle model the surface pulls light inwards, adding to the velocity component along the normal, so light should be faster in water. The wave model predicts slower, and Foucault measured slower.

10.2 Huygens principle

Imagine a small loudspeaker playing a steady note in the middle of a large empty hall. At one instant, mark every point where the air pressure has just reached its peak. These points lie on a sphere around the speaker: they are all the same distance away, so the disturbance reached them together and they oscillate in step.

A surface like this, on which every point is at the same stage (phase) of its oscillation, is called a wavefront (तरंगाग्र). As time passes the wavefront moves outward at the wave speed v, and the wave carries its energy at right angles to it. That is why rays, the lines along which energy flows, are drawn perpendicular to the wavefronts in any medium in which light has the same speed in every direction.

Definition
Wavefront

A surface on which every point of a wave oscillates in the same phase (a surface of constant phase). It moves with the wave speed, and rays are perpendicular to it.

ExampleAround a small lamp, the wavefronts are spheres centred on the lamp.

The shape of a wavefront depends on the source and on how far away you are (Fig. 10.2). A point source sends out spherical wavefronts, with rays spreading out radially. Far from the source, any patch you look at is a tiny piece of an enormous sphere and is practically flat: a plane wavefront, with parallel rays. Sunlight at your window is an example. The Sun is 1.5×1011 m away, so across a window 1 m wide its wavefront bulges by only about 8×10−13 m, roughly a millionth of a wavelength. For a bulb 2 m from the same window the bulge is about 6 cm, so its wavefronts are clearly curved.

Fig. 10.2Near a point source the wavefronts are spheres and the rays spread out; far away, a small patch of wavefront is practically a plane and the rays are parallel.NCERT Fig. 10.1 · p. 257

Two panels separated by a dashed line. Left, labelled near the source: a small yellow source with five blue circular arcs around it, like ripples, and five orange arrows (rays) spreading out radially from the source. Right, labelled far away: five blue lines that are almost straight and parallel, curving only very slightly, with three orange rays running parallel across them; the label says almost planes.

Huygens' insight was a recipe for finding where a wavefront will be a little later, if we know where it is now: treat every point on the present wavefront as a small source in its own right.

PrincipleHuygens' principle

Every point on a wavefront acts as a source of secondary wavelets that spread out with the speed of the wave. After a time τ, the new wavefront is the surface that touches all these wavelets on the forward side: their forward envelope.

Fig. 10.3 applies the recipe to part of a spherical wavefront. Around each marked point, draw a sphere of radius vτ. The surface that touches all of them on the outside is again a sphere centred on the source, larger by vτ: the wave has simply moved outward. For a plane wavefront the same construction gives a parallel plane a distance vτ further on. In both cases the rays, which are normals to the wavefronts, carry on in straight lines, as they should in a uniform medium.

3/4The surface touching all the wavelets in front is the wavefront at time τ.

Fig. 10.3Wavelets from each point of a wavefront, and the envelope that becomes the new wavefront.NCERT Fig. 10.2 · p. 257

A yellow source sits at the bottom centre. Step 1: a blue arc above it is the wavefront at time zero, with five dots marked on it. Step 2: a grey circle of radius v tau is drawn around each dot, labelled wavelets. Step 3: a thick orange arc touching all the circles on the outside appears, the wavefront at time tau, with five black arrows (rays) running outward from the dots to it. Step 4: a dashed vermilion arc touching the circles on the inside, near the source, is added and labelled no backwave.

Quick check
1 of 3
A long, straight tube light glows evenly along its length. Close to it, what is the shape of its wavefronts?
Show answer

Answer (C) Cylinders with the tube along their axis

Every point at the same distance from the line of the tube is reached at the same time, so the surfaces of constant phase are cylinders around it. Spheres would need a point source, and planes a very distant one.

10.3 Refraction and reflection of plane waves using Huygens principle

Huygens' principle earns its keep when a wavefront meets the boundary between two media. The part of the front that has crossed travels at a different speed from the part that has not, and the construction shows exactly how the front turns.

10.3.1 Refraction of a plane wave

Let light travel at speed v1 in medium 1 and v2 in medium 2, with v2<v1 (from air into glass, say). A plane wavefront meets the flat boundary at an angle, so one end of it touches the boundary before the other (Fig. 10.4).

DerivationSnell's law from wavelets

Call the end of the wavefront that has just reached the boundary P, and its other end Q. Light from Q still has to travel a distance QR in medium 1 to reach the boundary at R. Let this take a time τ, so QR=v1τ. During the same time, the wavelet that started from P grows in medium 2 to a radius v2τ.

By Huygens' principle, when light from Q arrives at R, the refracted wavefront is the plane through R that touches this wavelet. Call the point of contact S. Then PS=v2τ, and PS is perpendicular to RS, since a radius meets a tangent at right angles.

The angle between a wavefront and the boundary equals the angle between the ray and the normal, because each line in the second pair is turned through a right angle from its partner in the first. So angle QPR is the angle of incidence i, and angle PRS is the angle of refraction r. The right-angled triangles PQR and RSP share the hypotenuse PR, so

sin⁡i=QRPR=v1τPR
(10.1)
sin⁡r=PSPR=v2τPR
(10.2)

Dividing one by the other, the unknowns τ and PR cancel:

sin⁡isin⁡r=v1v2
(10.3)

4/4Right-angled triangles with the common side PR give (10.1) and (10.2).

Fig. 10.4In the time light from Q reaches R, the wavelet from P grows only to v2τ in the slower glass, so the front turns.NCERT Fig. 10.4 · p. 258

The boundary is a vertical line, with air on the left and shaded glass on the right. Step 1: three parallel orange rays come in from the upper left; a thick blue wavefront PQ is perpendicular to them, with P on the boundary and Q further back; a dashed blue arrow shows the path QR, of length v1 tau, that light from Q must still cover to reach the boundary at R. Step 2: a dashed orange semicircle of radius v2 tau is drawn in the glass around P, smaller than QR. Step 3: a thick orange line from R touching the semicircle at S is the refracted wavefront; the rays continue in the glass bent towards the normal, and the angles i at P and r at R are marked. Step 4: the triangles PQR and RSP are shaded and the ratios sin i = v1 tau over PR and sin r = v2 tau over PR are shown.

Equation (10.3) is the law of refraction, and it carries a prediction. A ray that bends towards the normal has r<i, so sin⁡i/sin⁡r>1 and therefore v2<v1: light must travel more slowly in the medium in which it bends towards the normal. This is the wave model's side of Fig. 10.1, and Foucault's measurement agrees with it.

The refractive index of a medium compares the speed of light c in vacuum with its speed in the medium:

n1=cv1
(10.4)
n2=cv2
(10.5)

Substituting v1=c/n1 and v2=c/n2 into (10.3) gives Snell's law in its familiar form:

Snell's law
n1sin⁡i=n2sin⁡r
(10.6)

What happens to the wavelength? Choose τ to be exactly one period T of the wave. In one period each part of the wave advances by one wavelength, so QR=λ1 and PS=λ2, and (10.1) and (10.2) give

λ1λ2=v1v2orv1λ1=v2λ2
(10.7)

Since v/λ is the frequency ν, the frequency is the same on both sides of the boundary. Only the speed and the wavelength change, and both change by the same factor n1/n2. Light entering glass of refractive index 1.5 from air keeps its frequency, but its speed and its wavelength drop to two-thirds of their values in air.

Example

A laser beam entering glass

A green laser pointer (λ=532 nm in air) shines on a glass block of refractive index 1.52 at an angle of incidence of 50∘. Find (a) the angle of refraction, (b) the speed of light in the glass, and (c) the frequency and the wavelength of the light in the glass. Take c=3.00×108 m/s and treat air as vacuum.

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

10.3.2 Refraction at a rarer medium

Now reverse the situation, so that light passes from glass or water into air and v2>v1. The construction is the same, but the wavelet in medium 2 now grows faster than light from Q closes the gap to the boundary. The refracted wavefront tilts the other way and the ray bends away from the normal, r>i, with Snell's law still holding (Fig. 10.5).

As i increases, r reaches 90∘ first. The angle of incidence at which this happens is the critical angle ic (क्रांतिक कोण). Putting r=90∘ in (10.6) gives

Critical angle
sin⁡ic=n2n1
(10.8)

The construction also shows why nothing can be refracted beyond this angle. By (10.2), at i=ic the wavelet from P grows, in the time τ, just far enough to reach R itself: v2τ=PR, and the refracted wavefront stands perpendicular to the boundary, skimming along it. For i>ic the wavelet has already passed R by the time light from Q arrives there, and no plane through a point inside a sphere can touch the sphere. So there is no refracted wave at all, and all the light is reflected back into the denser medium. This is the total internal reflection you met in the previous chapter.

2/3i=ic=41.8∘: the refracted ray skims along the surface (r=90∘).

Fig. 10.5From glass into air: bending away from the normal, grazing at the critical angle, then total internal reflection. Drawn for glass of index 1.5, so ic=41.8∘.NCERT Fig. 10.5 · p. 260

Glass fills the lower part of the picture and air the upper part, with a dashed normal at the point where a ray meets the surface. Step 1: a ray inside the glass meets the surface at a small angle of incidence i and leaves into the air at a larger angle r, bent away from the normal, with a faint dashed reflected ray. Step 2: the ray meets the surface at the critical angle, and the refracted ray skims along the surface at 90 degrees; the label gives sine of i c equals n2 over n1. Step 3: the angle of incidence is larger still; there is no ray in the air, and a strong reflected ray goes back down into the glass, labelled total internal reflection.

Example

Why a diamond sparkles

The refractive index of diamond is 2.42. (a) Find the critical angle for light trying to leave a diamond into air. (b) Inside a cut diamond, light meets a face at an angle of incidence of 35∘. What happens to it? (c) Would the same light escape from glass of index 1.50?

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

10.3.3 Reflection of a plane wave by a plane surface

DerivationThe law of reflection from wavelets

Let a plane wavefront PQ meet a plane mirror at an angle i (Fig. 10.6). End P touches the mirror first. Light from Q needs a time τ to reach the mirror at R, so QR=vτ. Meanwhile the wavelet from P spreads back into the same medium and reaches the same radius vτ.

The reflected wavefront is the plane through R that touches this wavelet, say at S. Triangles PQR and RSP each have a right angle (at Q and at S), share the hypotenuse PR, and have QR=PS=vτ. They are therefore congruent, so angle QPR equals angle SRP: the incident and reflected wavefronts make equal angles with the mirror. These angles equal the angles that the rays make with the normal, so i=r, which is the law of reflection.

3/3The tangent RS is the reflected wavefront; the congruent triangles give i=r.

Fig. 10.6The wavelet from P grows to the same radius vτ that light from Q travels, so the two shaded triangles are congruent and i=r.NCERT Fig. 10.6 · p. 261

A vertical mirror, hatched on its right-hand side, stands in the middle of the picture. Step 1: three parallel orange rays come in from the upper left towards the mirror; a thick blue wavefront PQ crosses them, with P on the mirror and Q further back, and a dashed blue arrow from Q to R on the mirror shows the distance v tau that light from Q still has to travel. Step 2: a dashed orange arc of radius v tau is drawn around P on the mirror's front side. Step 3: a thick orange line from R touching the arc at S is the reflected wavefront; reflected rays head back down to the left, the triangles PQR and RSP are shaded, equal angles i and r are marked, and a note states that the triangles are congruent, so i equals r.

With reflection and refraction explained, we can follow a wavefront through any optical device. One rule covers them all: a part of the front that spends longer in slow glass, or that has further to travel before a mirror sends it back, falls behind the rest, and the front changes shape accordingly (Fig. 10.7, Table 10.1). Because rays are perpendicular to the front, its new shape tells us at once where the rays go.

Table 10.1What optical devices do to a plane wavefront
Part of the front held back mostFront that emergesRays
Thin prismthe part crossing the thick basea tilted planeturned towards the base
Convex lensthe centre (thickest glass)a sphere shrinking onto the focusconverge
Concave lensthe edges (thickest glass)a sphere spreading from a point behind the lensdiverge
Concave mirrorthe centre (the edges are reflected first)a sphere shrinking onto the focusconverge
Convex mirrorthe edges (the centre is reflected first)a sphere spreading from a point behind the mirrordiverge

2/4The thick centre delays the middle: the front converges to the focus F.

Fig. 10.7Wherever the wave spends longer in glass, or reaches a mirror sooner, that part of the front changes its timing, and the front changes shape.NCERT Fig. 10.7 · p. 261

Plane blue wavefronts arrive from the left in every step. Step 1, prism: a triangular glass prism with its base at the bottom; the orange wavefronts leaving it are tilted, top ahead, and an arrow shows the light turned down towards the base. Step 2, convex lens: orange arcs after the lens are hollow towards the right and shrink onto a focus F. Step 3, concave lens: orange arcs after the lens bulge to the right and grow, centred on a hollow point F in front of the lens. Step 4, concave mirror: a curved mirror on the right sends back orange arcs that are hollow towards a focus F in front of it, with an arrow showing the reflected light travelling back to the left.

Example 10.1NCERT Example 10.1 · p. 262

Frequency, speed and energy at a boundary

A green laser beam (532 nm in air) is aimed at a swimming pool. A swimmer under water sees a green spot where it enters, and some of the beam is reflected towards a spectator.

(a) The swimmer and the spectator both see green. What fixes the frequency of the light inside the water and in the reflected beam, and why is it the same as the frequency of the incident light? (b) Under water the beam travels at about 2.25×108 m/s, yet after it leaves the pool through a glass side wall it is back to 3.00×108 m/s. Where does the energy to speed it up again come from? (c) A second laser of the same colour is ten times brighter. Compare the energy of one of its photons, and the number of photons it delivers each second, with those of the first laser.

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

Quick check
1 of 4
Orange light of wavelength 600 nm in air enters glass of refractive index 1.50. What is its wavelength in the glass, in nanometres?
nm
Show answer

Answer 400 nm

λ=λ0/n=600/1.50=400 nm. The frequency is unchanged, so the light is still orange.

10.4 Coherent and incoherent addition of waves

What happens when two waves reach the same point? For ripples, sound and light alike, experiment gives a simple answer, the superposition principle (अध्यारोपण का सिद्धांत):

PrinciplePrinciple of superposition

Where waves overlap, each one disturbs the medium exactly as it would if the others were absent, and the actual disturbance at any point and instant is the vector sum of these separate disturbances.

For light, the 'displacement' is the electric field of the wave. Interference is this principle applied to two waves of the same frequency, and whether we see anything interesting depends on whether the two waves keep in step.

Fix two prongs to one vibrating arm so that they dip into a ripple tank together. Each prong is a source of circular ripples, and because a single motor drives both, they rise and fall in step indefinitely. Sources whose phase difference stays constant like this are called coherent sources (कला-संबद्ध स्रोत). Fig. 10.8 shows a snapshot of their ripples: crests as solid circles and troughs as dashed ones.

Along some lines crest always meets crest and trough meets trough, and the water heaves twice as high as with one source. Along others crest always meets trough, and the water hardly moves. Which happens at a point P depends only on the path difference Δ=S2P−S1P (पथांतर), measured in wavelengths.

If Δ=0, as anywhere on the perpendicular bisector of S1S2, the two waves set off in step and arrive in step. If Δ=λ or 3λ, the wave from the farther source arrives one or three whole cycles late, which puts it in step again. But if Δ=λ/2 or 3.5λ, the late wave is behind by an odd number of half cycles, so crest arrives with trough. A path difference Δ corresponds to a phase difference ϕ=2πΔ/λ. With n=0,1,2,…, the two conditions are

Constructive interference
∣S2P−S1P∣=nλ
(10.9)
Destructive interference
∣S2P−S1P∣=(n+12)λ
(10.10)

Each condition picks out a curve. Points with a fixed difference of distances from two fixed points lie on a hyperbola, so the lines of strong motion (antinodal lines) and of stillness (nodal lines) in Fig. 10.8 are hyperbolas. They fan out from between the sources and become nearly straight far away, and a screen placed across them shows a row of strong and weak spots.

2/3Green: Δ=nλ, crest meets crest. Dashed: Δ=(n+12)λ, crest meets trough.

Fig. 10.8Two sources driven in step. Strong motion where the path difference is a whole number of wavelengths, stillness where it is an odd number of half wavelengths. Here S1S2=5λ.NCERT Fig. 10.8 · p. 262

Two yellow sources S1 and S2 sit one above the other on the left. Step 1: around each, solid blue circles mark crests and dashed grey circles mark troughs, overlapping to the right. Step 2: the circles fade and thick green curves (antinodal lines, where crest meets crest) fan out from between the sources, labelled path difference 0 and lambda; dashed vermilion curves between them (nodal lines, where crest meets trough) are labelled lambda over 2. Step 3: a vertical screen on the right has a graph of intensity beside it, bulging out wherever a green line meets the screen and touching the screen where a dashed line meets it.

DerivationIntensity for any phase difference

Between these extremes the intensity varies smoothly. Let each source on its own produce an amplitude a at P; the sources are far away compared with their separation, so the two amplitudes are equal. The displacements at P are

y1=acos⁡ωt
y2=acos⁡(ωt+ϕ)

Adding them with the identity cos⁡α+cos⁡β=2cos⁡α−β2cos⁡α+β2 gives

y=Acos⁡ ⁣(ωt+ϕ2)
A=2acos⁡ϕ2

This is again an oscillation at angular frequency ω, now with amplitude A=2acos⁡(ϕ/2). The intensity of a wave goes as its amplitude squared, so if one source alone gives I0∝a2,

I=4I0cos⁡2ϕ2
(10.11)

For ϕ=0,±2π,±4π,… (path difference nλ) this gives 4I0, four times one source and not twice. For ϕ=±π,±3π,… (path difference (n+12)λ) it gives zero (Fig. 10.9).

1/3In step: amplitude 2a, intensity 4I0.

Fig. 10.9Amplitudes add; intensity goes as amplitude squared. In step: 4I0. A quarter cycle apart: 2I0. Half a cycle apart: nothing.NCERT Eq. (10.11) · p. 264

Top: a solid blue wave y1 and a dashed orange wave y2 of the same amplitude. Bottom: their sum, a thick black wave. Right: a yellow bar shows the intensity. Step 1, phase difference zero: the two waves lie on top of each other, the sum has twice the amplitude, and the bar is full, 4 I nought. Step 2, phase difference pi over 2: the orange wave slides a quarter wavelength along, the sum has 1.41 times the amplitude, and the bar drops to half, 2 I nought. Step 3, phase difference pi: the orange wave is a half wavelength along, crest opposite trough, the sum is a flat line and the bar is empty.

Now suppose the two prongs are driven by separate motors that drift in and out of step. The phase difference ϕ at any point then wanders through all values, and the pattern of Fig. 10.10 slides back and forth across the tank. If ϕ changes much faster than we can observe, we see only the time average of cos⁡2(ϕ/2), which is 12:

Iˉ=4I0×12=2I0
(10.12)

Such sources are incoherent: their intensities simply add, and no fringes appear. The two headlights of a car shining on a garage door behave like this. Where both beams fall, the door is twice as bright as with one, evenly, without a trace of a pattern.

3/3Rapid random jumps average out to a uniform 2I0.

Fig. 10.10If the phase difference stays fixed the fringes stand still. If it jumps about, the fringes slide to and fro, and the eye sees only their average, 2I0 everywhere.NCERT Eq. (10.12) · p. 264

A graph of intensity against position on a screen, with dashed reference lines at 4 I nought and 2 I nought. Step 1: a blue curve rises and falls between 0 and 4 I nought, labelled steady phase difference. Step 2: the earlier curve is shown faint and dashed, and an orange curve of the same shape is shifted sideways, labelled phase jumps: fringes slide. Step 3: several faint shifted curves lie on top of each other and a thick green horizontal line at 2 I nought shows their average, labelled the eye sees the average.

Example

Two loudspeakers, one tone

Two loudspeakers side by side are driven in phase by the same amplifier with a 680 Hz tone, and sound travels at 340 m/s. Assume that each speaker alone gives the same intensity I0 at every seat considered. (In reality the nearer speaker is a little louder, so a quiet spot is not perfectly silent.) Find the intensity at a seat that is (a) 3.00 m from one speaker and 3.75 m from the other, (b) 3.00 m and 4.00 m away, (c) 3.000 m and 3.125 m away.

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

Quick check
1 of 4
Two coherent sources each give intensity I0 at a point where the path difference is λ/6. What is the resultant intensity, as a multiple of I0?
× I₀
Show answer

Answer 3 × I₀

ϕ=2π(λ/6)/λ=π/3, so I=4I0cos⁡2(π/6)=4I0×34=3I0.

10.5 Interference of light waves and Young's experiment

Try the ripple-tank experiment with light: put a small green LED behind each of two pinholes and look at a screen beyond them. It is evenly lit, with no fringes. The reason lies in how such sources make light. The light is a jumble of short wave trains from countless atoms, and its phase wanders at random; for a green LED it is effectively scrambled every few times 10−14 s. In the twentieth of a second your eye needs to register a picture, any fringe pattern would be replaced more than a million million times. Two independent sources are always incoherent.

Young's way round this was not to look for better lamps but to split one source in two. In Fig. 10.11 a single narrow slit S passes light from a small region of the lamp, and that light reaches two slits S1 and S2 a fraction of a millimetre apart. The two slits are now like two loudspeakers wired to the same amplifier: however erratically the lamp behaves, both receive the same wave, so the difference between their phases never changes. Today a laser, whose light stays in step over long distances, makes the experiment easy. Beyond the slits, bright and dark bands called interference fringes (व्यतिकरण) appear on a screen.

4/4Bright where the extra path is nλ; they are spaced β=λD/d apart.

Fig. 10.11One slit feeds two, so the two slits are coherent. The extra path to P is dsin⁡θ≈xd/D, and evenly spaced fringes appear on the screen.NCERT Fig. 10.12 · p. 265

From left to right: a yellow lamp, a barrier with one slit S, a barrier with two slits S1 and S2 a distance d apart, and a screen a distance D away. Step 1: blue arcs spread from S to the two slits and from each slit towards the screen. Step 2: a blue line from S1 and an orange line from S2 meet at a point P on the screen, a height x above the central point O, with a dashed line from the slits to O. Step 3: an enlarged inset shows the two slits with parallel rays leaving at angle theta; a dashed line from S1 meets the lower ray, cutting off a short thick vermilion segment, the path difference d sine theta, approximately x d over D. Step 4: a strip beside the screen shows evenly spaced yellow bright fringes with dark gaps, and the spacing beta between neighbouring bright fringes is marked.

DerivationWhere the fringes fall

Let the slits be a distance d apart and the screen a distance D away, with D≫d. Point O on the screen is equally far from both slits. Take a point P a height x above O. By Pythagoras, (S2P)2=D2+(x+d/2)2 and (S1P)2=D2+(x−d/2)2. Subtracting, the D2 terms cancel and (S2P)2−(S1P)2=2xd.

The left side factorises as (S2P−S1P)(S2P+S1P), and when x and d are much smaller than D, S2P+S1P≈2D. So the path difference Δ=S2P−S1P is

Δ≈xdD

Equivalently Δ=dsin⁡θ, where θ is the angle at which P is seen from the slits, since sin⁡θ≈tan⁡θ=x/D for small angles. Bright fringes need Δ=nλ and dark fringes Δ=(n+12)λ, so for n=0,±1,±2,…

Bright fringes
xn=n λDd
(10.13)
Dark fringes
xn=(n+12)λDd
(10.14)

Neighbouring bright fringes, and neighbouring dark ones, are separated by the same distance, the fringe width:

Fringe width
β=xn+1−xn=λDd

Four things follow from (10.13) and (10.14):

  • The fringes are equally spaced, bright and dark alternating, with a bright fringe at the centre O, where the path difference is zero.
  • β∝λ: red fringes are wider than blue ones. In white light the central fringe is white, since every colour has zero path difference there, and it is flanked by a few coloured fringes that soon blur together.
  • β∝1/d: bringing the slits closer spreads the fringes out. That is why the slits must be very close, typically a fraction of a millimetre apart.
  • Immersing the whole apparatus in a liquid of refractive index n shortens the wavelength to λ/n, so the fringes close up by the same factor.
InteractiveYoung's double-slit lab

A simulated screen shows the fringes from two slits in the colour of the chosen light, with a graph of intensity against position from minus 15 to plus 15 millimetres below it. Sliders set the wavelength from 400 to 700 nanometres, the slit separation from 0.10 to 1.00 millimetre and the screen distance from 0.5 to 3.0 metres, and a menu sets the width of each slit (ideal, 0.02, 0.05 or 0.10 millimetre), which adds a single-slit envelope. A switch replaces the coherent slits with two separate lamps, which gives an even 2 I nought, and a button overlays snapshots of the shifting pattern. A probe slider picks a point on the screen and reports its path difference in nanometres and in wavelengths, the phase difference and the intensity. Readouts give the fringe width and the angular fringe width.

Fig. 10.12Change the colour, the slit separation and the screen distance, and watch the fringe width. Then give the slits a width, or switch to two separate lamps.
Example

Fringes from a green laser

Light from a green laser (λ=532 nm) falls on two slits 0.40 mm apart, and the fringes are seen on a wall 1.6 m away. Find (a) the fringe width, (b) the distance from the central bright fringe to the third dark fringe, (c) the angular fringe width, and (d) the fringe width if the whole apparatus is immersed in water (n=1.33).

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

Quick check
1 of 4
Slits 0.50 mm apart are lit with light of wavelength 600 nm, and the screen is 1.5 m away. What is the fringe width, in millimetres?
mm
Show answer

Answer 1.8 mm

β=λD/d=(600×10−9)(1.5)/(5.0×10−4)=1.8×10−3 m=1.8 mm.

10.6 Diffraction

Remember the dark lines you saw between two nearly touching fingers held up to a tube light (Section 10.1)? Light squeezing through the narrow gap does not go straight on as ray optics says. It spreads out, and the spread-out light forms bright and dark bands. With a small enough source you can see the same thing at the edge of any shadow, which is never perfectly sharp. This spreading of waves around obstacles and through openings is called diffraction (विवर्तन).

Every kind of wave diffracts. Sea waves passing the end of a harbour wall curl round into the sheltered water behind it, and the band playing in the next street reaches you round the corner of a building. Light diffracts just as readily, but its wavelength is about a million times shorter than that of sound, so the effects become noticeable only for openings and obstacles a fraction of a millimetre across. That is why everyday shadows look sharp, and why ray optics works so well.

Diffraction also sets a hard limit on optical instruments. A lens or mirror of finite size cannot focus a point of light to a perfect point: it makes a tiny diffraction blur, and two objects closer together than about the size of that blur cannot be told apart, however well the lens is made. The finest detail a microscope or telescope can show is therefore set by the wavelength of the light. You can see diffraction by a regular pattern of openings, too: look at a distant street light through the cloth of an umbrella, and each gap between the threads spreads the light into a grid of spots fringed with colour.

10.6.1 The single slit

Let a parallel beam of monochromatic light fall straight onto a narrow slit of width a, and look at the light on a distant screen (or at the focus of a lens placed after the slit). What appears is not a sharp strip the width of the slit. The middle of the screen holds a wide bright band, and on each side lies a row of much dimmer bands, separated by dark gaps and fading with distance. The narrower the slit, the more the whole pattern spreads.

Treat the slit as a row of many tiny wavelet sources (Huygens again). The incoming wavefront lies flat across the slit, so every source starts at the same phase; what differs is how far each one is from the point where we look. In a direction θ, the wavelet from the lower edge travels asin⁡θ≈aθ further than the one from the upper edge, and the wavelets in between are spread evenly over this range (Fig. 10.13).

DerivationWhere the dark fringes fall

Straight ahead (θ=0): every wavelet travels the same distance, so they all arrive in phase and add up. This is the bright central maximum.

At aθ=λ: divide the slit into an upper half and a lower half. Each point in the upper half has a partner exactly a/2 below it, and the paths from the two partners differ by λ/2. Every pair cancels, so the intensity is zero: the first dark fringe.

At aθ=2λ: divide the slit into quarters. The first and second quarters cancel in the same way, and so do the third and fourth, so it is dark again. In general the minima are at

Single-slit minima
θ≈nλa
where
n
±1,±2,±3,… (not zero)

At aθ=32λ: divide the slit into thirds. Two thirds cancel, and the light from the remaining third survives, so there is a secondary maximum roughly midway between the first two minima. It is weak, since only a third of the slit contributes and its own wavelets are partly out of step. A full calculation gives about 4.7 % of the central intensity for the first secondary maximum and 1.6 % for the next. The secondary maxima lie near θ≈±(n+12)λ/a, with n=1,2,…

2/4The two halves cancel pair by pair: the first minimum.

Fig. 10.13Pair up the wavelets: whole halves, quarters, sixths… cancel to give the dark fringes. Angles are exaggerated.NCERT Fig. 10.14 · p. 267

Left: plane blue wavefronts arrive at a slit in a barrier; eight dots mark wavelet sources across the slit, and grey parallel rays leave them in one direction. A coloured bar beside the slit shows how it is divided. Right: a graph of intensity against angle, with a tall central peak and small side peaks, and a red dot marking the current direction. Step 1: the rays go straight ahead and the bar is one green block, all in step: bright; the dot is on top of the central peak. Step 2: the rays tilt so that the path difference between the edges, a thick red segment, is one wavelength; the bar is split into a blue upper half and an orange lower half that cancel: dark; the dot is at the first zero. Step 3: the tilt gives a path difference of one and a half wavelengths; the bar is in thirds, two cancel and one green third is left: faint; the dot is on the first small side peak. Step 4: the path difference is two wavelengths; the bar is in four alternating quarters that cancel in pairs: dark; the dot is at the second zero.

The central maximum runs from −λ/a to +λ/a, so its angular width is 2λ/a, twice that of the other bands. On a screen at a distance D it is 2λD/a wide. Making the slit narrower spreads the pattern out; making it much wider than the wavelength shrinks the pattern to a sharp image of the slit, and ray optics takes over.

Example

The width of the central band

Light of wavelength 600 nm falls normally on a slit 0.10 mm wide, and the pattern is seen on a screen 2.0 m away. Find (a) the angle of the first minimum, (b) the width of the central maximum on the screen, and (c) that width if blue light of wavelength 450 nm is used instead.

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

Compare

Interference and diffraction

Interference (double slit)

  • Light from a few coherent sources (two slits) is added
  • Fringes all of the same width, λD/d
  • For very narrow slits, bright fringes of nearly equal brightness
  • Spacing set by the slit separation d

Diffraction (single slit)

  • Light from a continuous band of wavelets across one opening is added
  • Central band twice as wide as the others
  • Brightness falls off rapidly away from the centre
  • Spread set by the opening width a

Both effects are superposition at work, and there is no deep physical line between them. Richard Feynman remarked that the difference is only a matter of usage: when a few sources are combined we tend to say interference, and when very many are combined we say diffraction.

In a real double-slit experiment each slit has a width of its own, so both effects act at once. Each slit spreads its light into a single-slit pattern, and within that spread the two slits interfere. The result is Young's fringes, spaced λD/d, whose brightness follows the single-slit pattern as an envelope (Fig. 10.14). Where an interference maximum falls on a diffraction minimum, that fringe is missing.

Fig. 10.14A real double slit: interference fringes inside a single-slit envelope. Drawn for slits whose centres are four slit-widths apart, so the 4th-order fringes fall on the envelope's first zeros and are missing.NCERT 10.6.1 · p. 268

Three graphs of intensity against angle, one above the other, sharing the same horizontal scale. Top, one slit: a broad central peak with small side peaks. Middle, two ideal slits: a row of equally tall, equally spaced narrow peaks. Bottom, two real slits: the same narrow peaks, but their heights follow the shape of the top graph, drawn as a dashed envelope; the fourth peak on each side, which falls on a zero of the envelope, is missing and labelled so.

Example

Fringes inside the central envelope

Two slits, each 0.060 mm wide, have their centres 0.30 mm apart. How many bright interference fringes lie within the central diffraction maximum?

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

10.6.2 Seeing the single slit diffraction pattern

You can see a single-slit pattern with things found at home or in the school laboratory. All you need is a narrow, adjustable slit and a thin, bright line of light.

Now hold two fingers almost touching in front of the same lamp, as at the start of the chapter. The dark lines you see in the gap are this same diffraction pattern, made by a slit whose width you can change by squeezing your fingers together.

Here is a check worth making. For two slits, the intensity swings between 0 and 4I0, and its average over a fringe is 2I0: exactly what the two slits would deliver separately. A dark fringe is therefore not light destroyed but light sent elsewhere, into the neighbouring bright fringes. The same bookkeeping holds for a single slit, where the energy spread over the whole pattern equals the energy that passed through the opening, as conservation of energy requires.

Quick check
1 of 4
A single slit is made narrower. The central bright band on the screen
Show answer

Answer (A) becomes wider

Its angular half-width is λ/a, so a smaller a gives a wider band. (It also gets dimmer, because less light gets through, but its width grows.)

10.7 Polarisation

Tie one end of a long rope to a pole and shake the free end up and down regularly. A wave runs along the rope (call this the x direction), and each bit of rope moves only up and down, along y:

y(x,t)=asin⁡(kx−ωt)
(10.15)

Here a is the amplitude, ω=2πν is the angular frequency, and

k=2πλ
(10.16)

is the angular wave number. Watch any one bit of the rope: it moves up and down along a fixed line, always at right angles to the direction in which the wave travels, so this is a transverse wave. A transverse wave in which every particle keeps to one fixed line is linearly polarised (रैखिकतः ध्रुवित). Here the line is the y axis, so this is a y-polarised wave; because the whole rope stays in the x–y plane, such a wave is also called plane polarised. Shake the rope from side to side instead and you get a z-polarised wave:

z(x,t)=asin⁡(kx−ωt)
(10.17)

If you keep changing the direction of shaking at random, many times a second, the vibration is still at right angles to the rope but has no fixed direction: the wave is unpolarised. Now pass the rope through a vertical slot in a board (Fig. 10.15). An up-and-down wave goes through, a side-to-side wave is stopped, and an unpolarised wave comes out vertically polarised, with only its up-and-down part left.

4/4A polaroid passes only the component of 𝐄 along its pass-axis.

Fig. 10.15A slot passes only the part of a transverse vibration along it. A polaroid does the same for the electric field of light.NCERT Fig. 10.17 · p. 269

Three columns, each an end-on view looking along the direction of travel, joined by arrows: what arrives, the filter, and what leaves. Step 1: the arriving rope vibrates up and down (a purple vertical double arrow); the filter is a wooden board with a vertical slot; the leaving wave still vibrates up and down, labelled passes. Step 2: the arriving rope vibrates side to side (a horizontal double arrow); nothing leaves, labelled blocked. Step 3: the arriving vibration is random, shown as double arrows in six directions; only a shorter vertical double arrow leaves, labelled only up-down. Step 4: the rope is replaced by light; orange double arrows in six directions show the electric field of unpolarised light, the filter is a polaroid disc with horizontal chain lines and a dashed vertical pass-axis, and a single vertical orange arrow leaves, labelled polarised.

Light behaves in the same way, and that is our evidence that light waves are transverse. In a light wave the electric field 𝐄 is always perpendicular to the direction of travel, and the direction in which 𝐄 oscillates is called the direction of polarisation (ध्रुवण). Light from the Sun, a bulb or a candle is unpolarised: its atoms emit independently, so the direction of 𝐄 changes rapidly and at random, though always within the plane perpendicular to the ray.

A polaroid acts on light as the slot acts on the rope. Every polaroid has one special direction in its plane, its pass-axis: the component of 𝐄 along this direction goes through, and the component at right angles to it is absorbed. What makes the direction special is the material. A typical sheet is polyvinyl alcohol, stretched so that its long-chain molecules line up, and stained with iodine so that electrons can move along the chains. An electric field along the chains drives small currents in them and loses its energy; a field across the chains cannot, and passes. So the pass-axis is perpendicular to the chains. Edwin Land invented polarising sheets in 1929, and this stretched, iodine-stained kind has been in use since 1938.

Put one polaroid in front of a lamp. Of the randomly oriented 𝐄, only the component along the pass-axis gets through. Averaged over all directions, that is half the intensity, whichever way you turn the sheet, so rotating it changes nothing you can see. The light that emerges, though, is now linearly polarised along the pass-axis.

Add a second polaroid. If its pass-axis is parallel to the first one's, the polarised light goes straight through. Turn it, and the light dims; at 90∘ (crossed polaroids) almost nothing gets through. Fig. 10.16 shows why. Let the light leaving the first sheet have amplitude E0 and intensity I0, and let the second pass-axis make an angle θ with the first. Only the component E0cos⁡θ along the second axis is transmitted, and intensity goes as amplitude squared.

LawMalus' law

When linearly polarised light of intensity I0 falls on a polaroid whose pass-axis makes an angle θ with the direction of polarisation, the transmitted intensity is I0cos⁡2θ.

Malus' law
I=I0cos⁡2θ
(10.18)

2/3θ=60∘: the field passed is E0cos⁡60∘=E0/2, so I=I0/4.

Fig. 10.16Only the component E0cos⁡θ gets through, so the intensity is I0cos⁡2θ: a quarter at 60∘, nothing when crossed.NCERT Fig. 10.18 · p. 271

Left: the first polaroid, a purple disc with a vertical dashed pass-axis and an orange vertical double arrow for the electric field E nought of the light leaving it. Middle: the second polaroid with its own dashed pass-axis, the incoming field shown faintly, and a green double arrow for the part it lets through. Right: a yellow intensity bar. Step 1: the second axis is vertical, the whole field passes and the bar is full, I nought. Step 2: the second axis is turned 60 degrees; a dashed line projects the field onto it, the green arrow is half as long, E nought cos 60 degrees, and the bar falls to a quarter, I nought over 4. Step 3: the axis is horizontal, crossed with the first; no green arrow remains and the bar is empty.

Two sheets therefore make a dimmer: turning one of them sets the transmitted intensity anywhere between half the lamp's intensity (axes parallel) and zero (axes crossed). As the second sheet turns through a full circle, the light passes through two maxima, at θ=0∘ and 180∘, and two minima, at 90∘ and 270∘.

InteractivePolaroid lab

A lamp sends unpolarised light, drawn as a star of arrows, through up to three polaroid discs, each with its pass-axis drawn as a line, to a light meter. Between the sheets an arrow shows the direction and size of the electric field. Sliders turn the first sheet, the optional middle sheet and the last sheet, and check boxes add or remove the middle and last sheets. Readouts give the intensity after each sheet as a percentage of the lamp's, with the Malus' law calculation written out. A graph plots the transmitted intensity against the angle of the last sheet from 0 to 360 degrees, with a dot for the current setting. A Play button turns the last sheet slowly, and a preset button crosses the outer sheets.

Fig. 10.17Turn each sheet and read the intensity after it. Try one sheet alone, two crossed sheets, and then slip a third sheet between the crossed pair.
Example

Two polaroids

Unpolarised light of intensity 50 W/m2 falls on two polaroids whose pass-axes are at 35∘ to each other. (a) What intensity emerges? (b) To what angle must the second sheet be turned, relative to the first, to let out 10 W/m2?

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

Example 10.2NCERT Example 10.2 · p. 271

A spinning sheet between crossed polaroids

Two polaroids P1 and P3 are crossed, so on their own they let no light through. A third polaroid P2 between them is spun at a steady 5 revolutions per second; at t=0 its pass-axis is parallel to that of P1. Unpolarised light of intensity 80 W/m2 falls on P1.

(a) Find the intensity leaving P3 when the axis of P2 is at an angle θ to that of P1. (b) What is it at t=1/60 s? (c) How many times each second is the emerging light at its brightest, and how bright is it then? (d) What average intensity does a slow light meter read?

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

Quick check
1 of 4
Unpolarised light of intensity 100 W/m2 passes through two polaroids whose pass-axes are at 60∘ to each other. What intensity emerges, in W/m2?
W/m²
Show answer

Answer 12.5 W/m²

The first sheet passes 50 W/m2; the second passes 50cos⁡260∘=50×0.25=12.5 W/m2.
Recap

Wave Optics: key points and summary

  1. The particle and wave models of light predict opposite speeds in water; light is measured to be slower there, as the wave model says (Section 10.1).
  2. A wavefront is a surface of constant phase, and rays are perpendicular to it. A point source gives spherical wavefronts; a distant source gives practically plane ones (Section 10.2).
  3. Huygens' principle: every point of a wavefront is a source of secondary wavelets, and their forward envelope is the wavefront a time τ later (Principle).
  4. Huygens' construction gives Snell's law, n1sin⁡i=n2sin⁡r with n=c/v, and predicts that light is slower in the medium where it bends towards the normal (Section 10.3.1).
  5. At a boundary the frequency is unchanged, while the speed and the wavelength change by the same factor ((10.7)).
  6. Light entering a rarer medium bends away from the normal. Beyond the critical angle, sin⁡ic=n2/n1, no refracted wave can form and the light is totally reflected (Section 10.3.2).
  7. Reflection obeys i=r. Prisms, lenses and mirrors reshape wavefronts, and the travel time from object to image is the same along every ray (Section 10.3.3).
  8. Superposition: amplitudes add. Two coherent sources give I=4I0cos⁡2(ϕ/2), bright for path difference nλ and dark for (n+12)λ; incoherent sources give 2I0 everywhere (Section 10.4).
  9. In Young's experiment one slit feeds two coherent slits. Bright fringes lie at x=nλD/d, dark ones at (n+12)λD/d, equally spaced with fringe width β=λD/d (Section 10.5).
  10. For a single slit of width a, the central maximum has angular width 2λ/a, the minima lie at θ=nλ/a, and the secondary maxima are much weaker. A real double-slit pattern is fringes inside this envelope (Section 10.6.1).
  11. Diffraction limits the detail optical instruments can resolve. Interference and diffraction only redistribute energy (Section 10.6).
  12. Light is transverse and can be polarised. A polaroid passes the component of 𝐄 along its pass-axis: it halves unpolarised light, and polarised light obeys Malus' law, I=I0cos⁡2θ (Section 10.7).
Flash cards · 11

Key relations

1 / 11
Practice · 15 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises 10.1

    A helium–neon laser beam of wavelength 632.8 nm in air strikes the polished face of a glass block of refractive index 1.50. Part of it is reflected and part enters the glass. Take c=3.00×108 m/s.

    1. What are the wavelength, speed and frequency of the reflected light? Give the frequency in hertz.

      Hz
      Show answer

      Answer 474000000000000 Hz

      The reflected light is still in air, so λ=632.8 nm, v=3.00×108 m/s, and ν=c/λ=(3.00×108)/(632.8×10−9)=4.74×1014 Hz.
    2. What is the speed of the light inside the glass, in m/s?

      m/s
      Show answer

      Answer 200000000 m/s

      v=c/n=(3.00×108)/1.50=2.00×108 m/s.
    3. What are its frequency and its wavelength inside the glass? Give the wavelength in nanometres.

      nm
      Show answer

      Answer 421.9 nm

      The frequency is unchanged, 4.74×1014 Hz, so λ=v/ν=λ0/n=632.8/1.50=421.9 nm.
  2. Short answerEasyNCERT Exercises 10.2

    For each situation below, say whether the wavefront is flat, a sphere that is spreading out, a sphere that is shrinking onto a point, or a cylinder.

    1. Light close to a long, straight, glowing filament.

      Show answer

      Answer (D) a cylinder

      Points at the same distance from a line form a cylinder around it.
    2. Light from a small bulb at the focus of a torch's concave reflector, after reflection.

      Show answer

      Answer (A) a plane

      A point source at the focus of a concave mirror gives a parallel beam, whose wavefronts are planes.
    3. Light from a distant lamp just after it has passed through a convex lens.

      Show answer

      Answer (C) a converging sphere

      The lens delays the centre most, so the front is a sphere shrinking onto the focus.
    4. Sunlight falling on your school playground.

      Show answer

      Answer (A) a plane

      The Sun is so far away that any patch of its spherical wavefront the size of a playground is flat to far better than a wavelength.
  3. Short answerEasyNCERT Exercises 10.3

    Answer both parts. Take c=3.00×108 m/s.

    1. Diamond has a refractive index of 2.42. What is the speed of light in diamond, in m/s?

      m/s
      Show answer

      Answer 124000000 m/s

      v=c/n=(3.00×108)/2.42=1.24×108 m/s.
    2. White light passes through a glass prism. Do its orange and blue components cross the glass at the same speed? If not, which is faster, and how does this explain why the prism spreads white light into colours?

      Show solution

      No. The refractive index of glass is slightly larger for shorter wavelengths, so by v=c/n blue light travels more slowly than orange: orange is faster. The bending at each face depends on n, so blue is deviated more than orange and the colours fan out. This is dispersion.

  4. NumericalMediumNCERT Exercises 10.4

    In a double-slit experiment the slits are 0.50 mm apart and the screen is 1.2 m away. The distance between the second bright fringe on one side of the central fringe and the third bright fringe on the other side is 6.0 mm. Find the wavelength of the light, in nanometres.

    nm
    Show solution

    The bright fringes lie at xn=nβ. The fringes n=−2 and n=+3 are 5β apart, so 5β=6.0 mm and β=1.2 mm.

    From β=λD/d: λ=βd/D=(1.2×10−3 m)(5.0×10−4 m)/(1.2 m)=5.0×10−7 m=500 nm (blue-green light).

  5. NumericalMediumNCERT Exercises 10.5

    In a double-slit experiment with light of wavelength λ, the intensity at a point where the path difference is λ/4 is 12 units.

    1. What is the intensity at the centre of a bright fringe?

      units
      Show answer

      Answer 24 units

      At Δ=λ/4, ϕ=π/2 and I=4I0cos⁡2(π/4)=2I0=12, so I0=6. A bright fringe has 4I0=24 units.
    2. What is the intensity where the path difference is λ/6?

      units
      Show answer

      Answer 18 units

      ϕ=π/3, so I=4I0cos⁡2(π/6)=4×6×34=18 units.
    3. What is the intensity where the path difference is λ/2?

      units
      Show answer

      Answer 0 units

      ϕ=π, so I=4I0cos⁡2(π/2)=0: the centre of a dark fringe.
  6. NumericalHardNCERT Exercises 10.6

    A discharge lamp gives out two strong lines, red at 630 nm and violet at 420 nm. Its light falls on two slits 0.45 mm apart, and the fringes are viewed on a screen 1.5 m away.

    1. What is the fringe width of the red fringes, in millimetres?

      mm
      Show answer

      Answer 2.1 mm

      β1=λ1D/d=(630×10−9)(1.5)/(4.5×10−4)=2.1×10−3 m=2.1 mm.
    2. How far from the central fringe is the third dark fringe of the violet light, in millimetres?

      mm
      Show answer

      Answer 3.5 mm

      β2=(420×10−9)(1.5)/(4.5×10−4)=1.4 mm. The dark fringes lie at 12β2,32β2,52β2,…, so the third is at 2.5×1.4=3.5 mm.
    3. Away from the centre, what is the smallest distance at which a red bright fringe and a violet bright fringe fall together, in millimetres?

      mm
      Show answer

      Answer 4.2 mm

      Coincidence needs 630 n1=420 n2, that is 3n1=2n2. The smallest solution is n1=2, n2=3, at x=2β1=4.2 mm (and 3β2=4.2 mm too).
  7. Quick answerEasy

    A surface on which every point of a wave oscillates in the same phase is called a .

    Show answer

    Answer wavefront

    A wavefront is a surface of constant phase; rays are perpendicular to it.
  8. Quick answerMedium

    A double-slit experiment is moved from air into water (n=4/3), with nothing else changed. What happens to the fringe width?

    Show answer

    Answer (B) It shrinks to 3/4 of its value

    In water λ′=λ/n=34λ, and β=λD/d, so the fringe width also becomes 34 of its value in air.
  9. Quick answerMedium

    Light of wavelength 550 nm passes through a slit 0.11 mm wide onto a screen 1.5 m away. What is the width of the central bright band, in millimetres?

    mm
    Show answer

    Answer 15 mm

    w=2λD/a=2(5.5×10−7)(1.5)/(1.1×10−4)=1.5×10−2 m=15 mm.
  10. Quick answerEasy

    Put the steps of Huygens' construction in order.

    1. Take that surface as the wavefront at time τ
    2. Draw the surface that touches all the spheres on the forward side
    3. Mark points on the wavefront at time t=0
    4. Draw a sphere of radius vτ around each point
    Show answer
    Answer
    1. Mark points on the wavefront at time t=0
    2. Draw a sphere of radius vτ around each point
    3. Draw the surface that touches all the spheres on the forward side
    4. Take that surface as the wavefront at time τ
    Points on the old front become sources; their wavelets grow to vτ; the forward envelope is the new front.
  11. Quick answerMedium

    Match each observation with what it shows about light.

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

    Show answer
    Answer
    • Polarisation by a polaroid : Light waves are transverse
    • Fringes in Young's experiment : Amplitudes from coherent sources add
    • Spreading after a narrow slit : Waves bend around openings and edges
    • Foucault's measurement in water : Light is slower where it bends towards the normal
    Each observation is explained by one feature of the wave model.
  12. Quick answerHard

    Unpolarised light of intensity 64 W/m2 passes through three polaroids. The second one's pass-axis is at 30∘ to the first one's, and the third one's is at 90∘ to the first one's. What intensity emerges, in W/m2?

    W/m²
    Show answer

    Answer 6 W/m²

    After the first sheet 32 W/m2; after the second 32cos⁡230∘=24 W/m2; after the third 24cos⁡260∘=6.0 W/m2. (Check: 324sin⁡260∘=6.0.)
  13. Quick answerEasy

    Two separate green LEDs, each behind its own pinhole, produce a steady pattern of interference fringes on a screen.

    Show answer

    Answer False

    The two LEDs are incoherent: the phase of each wanders at random within a few times 10−14 s, so any pattern averages out to an even illumination.
  14. Long answerHard

    Using Huygens' construction, explain why light going from glass into air at an angle of incidence larger than the critical angle produces no refracted wave.

    Show solution

    Let a plane wavefront PQ in glass reach the boundary first at P. Light from Q reaches the boundary at R after a time τ, with QR=v1τ=PRsin⁡i. In that time the wavelet from P grows in air to radius v2τ, which is larger than v1τ.

    A refracted wavefront would be a plane through R touching this wavelet, which is possible only if R lies outside it, that is if v2τ≤PR. Since v2τ=(v2/v1)PRsin⁡i=(n1/n2)PRsin⁡i, the condition is sin⁡i≤n2/n1=sin⁡ic.

    For i>ic, R lies inside the wavelet, so no plane through R can touch it and no refracted wavefront exists. The wave is totally reflected, and the reflected wavefront is found exactly as for a mirror.

  15. Short answerMedium

    A student notices that the bright fringes in Young's experiment are four times as bright as the light from one slit alone, and concludes that two slits produce more energy than they receive. Explain the mistake.

    Show solution

    The bright fringes do reach 4I0, but the dark fringes between them get nothing. Since cos⁡2 averages to 12, the intensity averaged over a fringe is 4I0×12=2I0, exactly the sum of what the two slits send separately. Interference moves energy from the dark fringes into the bright ones; it does not create any.

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

Quick check: MCQs on Wave Optics

  1. 10.1 Introduction3 questions
  2. 10.2 Huygens principle3 questions
  3. 10.3.3 Reflection of a plane wave by a plane surface4 questions
  4. 10.4 Coherent and incoherent addition of waves4 questions
  5. 10.5 Interference of light waves and Young's experiment4 questions
  6. 10.6.2 Seeing the single slit diffraction pattern4 questions
  7. 10.7 Polarisation4 questions
Glossary · 14

Key terms (English–Hindi)

coherent sourcesकला-संबद्ध स्रोतkala-sambaddh srot
Sources of the same frequency whose phase difference stays constant. Incoherent sources have a randomly changing phase difference.First seen in the chapter →
critical angleक्रांतिक कोणkrantik kon
The angle of incidence in the denser medium for which the angle of refraction is 90∘: sin⁡ic=n2/n1.First seen in the chapter →
diffractionविवर्तनvivartan
The spreading of a wave around obstacles and through openings, noticeable when they are not much larger than the wavelength.First seen in the chapter →
fringe widthफ्रिंज चौड़ाईfringe chaudai
The distance between neighbouring bright (or dark) fringes; β=λD/d for a double slit.First seen in the chapter →
interferenceव्यतिकरणvyatikaran
The redistribution of intensity when coherent waves overlap: bright where they arrive in step, dark where they arrive half a cycle apart.First seen in the chapter →
linearly polarisedरैखिकतः ध्रुवितraikhikatah dhruvit
Light whose electric field oscillates along one fixed line; also called plane polarised.First seen in the chapter →
Malus' lawमैलस का नियमMailas ka niyam
I=I0cos⁡2θ for polarised light of intensity I0 meeting a polaroid whose axis is at θ to it.
path differenceपथांतरpathantar
The difference between the distances from two sources to a point; it gives the phase difference ϕ=2πΔ/λ.First seen in the chapter →
polarisationध्रुवणdhruvan
The direction in which the vibration of a transverse wave (for light, 𝐄) takes place.First seen in the chapter →
polaroidपोलेरॉइडpoleroid
A plastic sheet of aligned long-chain molecules that passes only the component of 𝐄 along its pass-axis.First seen in the chapter →
secondary waveletद्वितीयक तरंगिकाdvitiyak tarangika
A small spherical wave spreading from a point on a wavefront, in Huygens' construction.First seen in the chapter →
superposition principleअध्यारोपण का सिद्धांतadhyaropan ka siddhant
Overlapping waves produce a displacement equal to the vector sum of the displacements each would produce alone.First seen in the chapter →
unpolarised lightअध्रुवित प्रकाशadhruvit prakash
Light whose electric field direction changes rapidly and at random, within the plane perpendicular to the ray.First seen in the chapter →
wavefrontतरंगाग्रtarangagra
A surface on which every point of a wave oscillates in the same phase. Rays are perpendicular to it.First seen in the chapter →