NCERT Solutions for Class 12 Physics Chapter 10: Wave Optics

These Class 12 Physics Chapter 10 solutions cover Wave Optics from the NCERT textbook (Part II), updated for the session 2026–27. Every NCERT Exercise question (10.1–10.10) is reproduced verbatim and solved step by step, with all numericals worked out and verified to the correct units, plus extra practice, MCQs, Assertion–Reason questions, exam tips and FAQs to help you master Huygens’ principle, interference, diffraction and polarisation.

Class: 12 Subject: Physics Chapter: 10 Name: Wave Optics Exercises: 10.1–10.10 Session: 2026–27
On this page

Class 12 Physics Chapter 10 Wave Optics – Overview

Chapter 10, Wave Optics, treats light as a wave rather than a ray and uses this picture to explain phenomena that ray (geometrical) optics cannot. It begins with the historical contest between the corpuscular model (Descartes, Newton) and the wave model (Huygens), settled in favour of waves by Young’s 1801 interference experiment and Foucault’s 1850 measurement that light travels slower in water. The chapter introduces the wavefront and Huygens’ principle, then uses them to derive the laws of reflection and refraction (Snell’s law). It develops the superposition principle to explain constructive and destructive interference, analyses Young’s double-slit experiment and its fringe formula, describes single-slit diffraction, and finally explains polarisation (Malus’ law and Brewster’s angle) as proof that light is a transverse electromagnetic wave.

Key Concepts & Definitions

Wavefront: the locus of all points of a wave that oscillate in the same phase — a surface of constant phase. Energy travels perpendicular to the wavefront. A point source gives a spherical wavefront; far from the source it is effectively plane.

Huygens’ principle: every point on a wavefront acts as a source of secondary spherical wavelets that spread out with the wave speed; the new wavefront at a later time is the forward common tangent (envelope) of these wavelets.

Coherent sources: two sources whose phase difference at any point stays constant in time. Only coherent sources of the same frequency give a stable interference pattern.

Interference: redistribution of light energy when waves superpose — bright (constructive) where path difference = nλ, dark (destructive) where path difference = (n + ½)λ.

Diffraction: bending/spreading of waves around obstacles and apertures comparable to the wavelength, giving a central bright maximum flanked by weaker secondary maxima.

Polarisation: restriction of the vibrations of the electric field of a light wave to a single plane. It occurs only for transverse waves, so it proves light is transverse.

Important Formulas

Snell’s law (wave form): sin i / sin r = v1/v2 = n2/n1; refractive index n = c/v.

Speed–wavelength relation: v = νλ; on refraction frequency ν is unchanged, while v and λ change.

Resultant intensity (two coherent sources): I = 4I0 cos2(φ/2), where φ is the phase difference.

YDSE fringe positions: bright xn = nλD/d; dark xn = (n + ½)λD/d.

Fringe width: β = λD/d; angular fringe width θ = λ/d.

Single-slit diffraction: minima at a sinθ = nλ (n = ±1, ±2, …); half-angular width of central maximum ≈ λ/a.

Polarisation: Malus’ law I = I0 cos2θ; Brewster’s law n = tan iB.

Ray-optics validity (Fresnel distance): zF = a2/λ.

NCERT Exercises (10.1–10.10) — Solutions

10.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

SOLUTION Given: λair = 589 nm = 589 × 10−9 m, nwater = 1.33, c = 3.0 × 108 m s−1. Frequency of the incident light: ν = c/λ = (3.0 × 108) / (589 × 10−9) = 5.09 × 1014 Hz. (a) Reflected light: reflection occurs in the same medium (air), so wavelength, frequency and speed are unchanged: λ = 589 nm, ν = 5.09 × 1014 Hz, speed = c = 3.0 × 108 m s−1. (b) Refracted light: frequency does not change on refraction, so ν = 5.09 × 1014 Hz. Speed in water v = c/n = (3.0 × 108)/1.33 = 2.26 × 108 m s−1. Wavelength in water λ′ = v/ν = (2.26 × 108)/(5.09 × 1014) ≈ 444 nm (= λair/n).

10.2 What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.

SOLUTION (a) A point source emits energy uniformly in all directions, so the wavefront is a spherical (diverging) wavefront centred on the source. (b) A point source at the focus of a convex lens produces a parallel emergent beam, so the wavefront becomes a plane wavefront. (c) A star is enormously far away, so by the time its spherical wavefront reaches Earth its radius is huge; the small portion intercepted by the Earth is essentially plane.

10.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s–1) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

SOLUTION (a) v = c/n = (3.0 × 108)/1.5 = 2.0 × 108 m s−1. (b) No. The refractive index of glass — and hence the speed of light in it — depends on the colour (wavelength). Violet deviates more than red in a prism, so nviolet > nred, which means violet light travels slower in glass than red light.

10.4 In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

SOLUTION Given: d = 0.28 mm = 0.28 × 10−3 m, D = 1.4 m, distance of 4th bright fringe x4 = 1.2 cm = 1.2 × 10−2 m, n = 4. For a bright fringe, xn = nλD/d, so λ = xn d / (n D). λ = (1.2 × 10−2)(0.28 × 10−3) / (4 × 1.4) = (3.36 × 10−6)/5.6 = 6.0 × 10−7 m = 600 nm.

10.5 In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3 ?

SOLUTION Resultant intensity for two coherent waves: I = 4I0 cos2(φ/2), where φ = (2π/λ) × (path difference). At path difference λ: φ = 2π, so I = 4I0 cos2(π) = 4I0 = K. Hence I0 = K/4. At path difference λ/3: φ = (2π/λ)(λ/3) = 2π/3, so I′ = 4I0 cos2(π/3) = 4I0 × (½)2 = 4I0 × ¼ = I0. Therefore I′ = I0 = K/4.

10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

SOLUTION Take the standard arrangement of this NCERT problem: slit separation d = 2 mm = 2 × 10−3 m and screen distance D = 120 cm = 1.2 m, with λ1 = 650 nm and λ2 = 520 nm. (a) Distance of the n-th bright fringe: x = nλ1D/d, with n = 3.
x = (3 × 650 × 10−9 × 1.2)/(2 × 10−3) = (2.34 × 10−6)/(2 × 10−3) = 1.17 × 10−3 m = 1.17 mm. (b) Bright fringes coincide when n1λ1 = n2λ2 ⇒ n1/n2 = λ2/λ1 = 520/650 = 4/5. The smallest whole numbers are n1 = 4 and n2 = 5. Least distance = n1λ1D/d = (4 × 650 × 10−9 × 1.2)/(2 × 10−3) = (3.12 × 10−6)/(2 × 10−3) = 1.56 × 10−3 m = 1.56 mm.

10.7 In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

SOLUTION Angular fringe width in air: θair = λ/d = 0.2°. When the set-up is immersed in water, the wavelength becomes λ′ = λ/n, while d is unchanged. So the angular width becomes θwater = λ′/d = (λ/d)/n = θair/n. θwater = 0.2° / (4/3) = 0.2° × (3/4) = 0.15°. (The given 1 m and 600 nm are not needed because the answer depends only on the ratio θair/n.)

10.8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

SOLUTION By Brewster’s law, the reflected light is completely plane-polarised when tan iB = n, where iB is the Brewster (polarising) angle. tan iB = 1.5 ⇒ iB = tan−1(1.5) ≈ 56.3°.

10.9 Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

SOLUTION Reflection takes place in the same medium, so the wavelength and frequency are unchanged. Wavelength of reflected light = 5000 Å (= 5 × 10−7 m). Frequency ν = c/λ = (3.0 × 108)/(5 × 10−7) = 6 × 1014 Hz. The reflected ray is normal (perpendicular) to the incident ray when the angle between them is 90°. The angle between incident and reflected rays equals 2i (i = angle of incidence), so 2i = 90° ⇒ i = 45°.

10.10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

SOLUTION Ray optics is a good approximation up to the Fresnel distance zF = a2/λ, where a is the aperture and λ the wavelength. Given a = 4 mm = 4 × 10−3 m and λ = 400 nm = 4 × 10−7 m. zF = a2/λ = (4 × 10−3)2 / (4 × 10−7) = (16 × 10−6)/(4 × 10−7) = 40 m. Up to about 40 m the spreading due to diffraction is small compared with the aperture, so ray optics holds.

Extra Practice Questions

Short Answer Type Questions

Q1. Define a wavefront and state how the rays are related to it.

ANSWERA wavefront is the locus of all points of a medium that are vibrating in the same phase (a surface of constant phase). The direction of energy propagation, i.e. the ray, is always perpendicular (normal) to the wavefront.

Q2. Why are two independent sodium lamps not used to obtain a sustained interference pattern?

ANSWERTwo independent lamps emit light with abrupt, random phase changes (about every 10−10 s), so they are incoherent. Their phase difference at any point varies rapidly and randomly, so no stable maxima and minima form; the intensities simply add and we see uniform illumination.

Q3. State Malus’ law and write its mathematical form.

ANSWERWhen plane-polarised light of intensity I0 passes through an analyser whose pass-axis makes an angle θ with the plane of polarisation, the transmitted intensity is I = I0 cos2θ. This is Malus’ law.

Q4. How does the fringe width in Young’s double-slit experiment change when the whole apparatus is immersed in a liquid of refractive index n?

ANSWERIn a medium of refractive index n the wavelength reduces to λ/n. Since β = λD/d, the fringe width also reduces by the factor n, i.e. β′ = β/n. The fringes therefore come closer together.

Q5. State Brewster’s law and define the polarising angle.

ANSWERBrewster’s law states that when unpolarised light is incident at the polarising angle iB, the reflected light is completely plane-polarised and tan iB = n (refractive index). The polarising (Brewster) angle is the angle of incidence at which the reflected and refracted rays are perpendicular to each other.

Long Answer Type Questions

Q1. State Huygens’ principle and use it to derive Snell’s law of refraction for a plane wave.

ANSWERHuygens’ principle: every point on a wavefront is a source of secondary wavelets that spread out in all directions with the speed of the wave; the new wavefront is the forward common tangent (envelope) of these wavelets. Consider a plane wavefront AB incident at angle i on the boundary between medium 1 (speed v1) and medium 2 (speed v2). While the wavelet from B travels BC = v1t in medium 1, the wavelet from A travels AE = v2t in medium 2. In triangles ABC and AEC, sin i = BC/AC = v1t/AC and sin r = AE/AC = v2t/AC. Dividing, sin i/sin r = v1/v2 = n2/n1 (since n = c/v), giving n1 sin i = n2 sin r, which is Snell’s law. Because r < i implies v2 < v1, the wave model correctly predicts that light slows down in a denser medium — opposite to the corpuscular model and confirmed by experiment.

Q2. Describe Young’s double-slit experiment and obtain expressions for the positions of bright and dark fringes and the fringe width.

ANSWERLight from a single source S passes through two close, parallel slits S1 and S2, which act as coherent sources because they are derived from the same wavefront. On a screen at distance D, the waves superpose. For a point at distance x from the centre, the path difference is approximately xd/D, where d is the slit separation. Constructive interference (bright fringe) occurs when xd/D = nλ, giving xn = nλD/d. Destructive interference (dark fringe) occurs when xd/D = (n + ½)λ, giving xn = (n + ½)λD/d. The separation between consecutive bright (or dark) fringes is the fringe width β = λD/d. The fringes are bright and dark bands, equally spaced and parallel to the slits, confirming the wave nature of light.

Q3. Explain single-slit diffraction qualitatively and write the condition for the minima. How does it differ from interference?

ANSWERWhen a parallel monochromatic beam falls on a single slit of width a, each part of the wavefront in the slit acts as a secondary source. At the centre (θ = 0) all contributions are in phase, giving a broad central maximum. Away from the centre, the slit can be imagined to split into pairs of strips whose contributions cancel; this gives minima (zero intensity) when a sinθ = nλ (n = ±1, ±2, …). Between minima there are weaker secondary maxima near a sinθ = (n + ½)λ. Differences from interference: interference (e.g. two slits) gives many equally spaced fringes of nearly equal brightness arising from a few discrete sources, whereas diffraction (one slit) gives a wide central maximum with rapidly fading side maxima arising from a continuous distribution of secondary sources. Feynman noted there is no sharp physical distinction — “interference” is used for few sources and “diffraction” for many.

Multiple Choice Questions (MCQs)

1. A wavefront is a surface on which all points have the same:

(a) amplitude only    (b) phase    (c) frequency only    (d) speed only

2. According to Huygens’ principle, each point on a wavefront acts as a source of:

(a) primary rays    (b) secondary wavelets    (c) photons    (d) electrons

3. When light passes from air into water, which quantity remains unchanged?

(a) speed    (b) wavelength    (c) frequency    (d) refractive index

4. In Young’s double-slit experiment the fringe width β is given by:

(a) λd/D    (b) λD/d    (c) dD/λ    (d) D/(λd)

5. For constructive interference, the path difference between two coherent waves must be:

(a) (n + ½)λ    (b) nλ    (c) (2n + 1)λ/4    (d) nλ/2

6. The maximum resultant intensity in two-source interference (each source intensity I0) is:

(a) I0    (b) 2I0    (c) 4I0    (d) 8I0

7. In single-slit diffraction, the first minimum occurs at an angle θ where:

(a) a sinθ = λ/2    (b) a sinθ = λ    (c) a sinθ = 2λ    (d) a sinθ = 3λ/2

8. Polarisation of light proves that light waves are:

(a) longitudinal    (b) transverse    (c) mechanical    (d) stationary

9. According to Malus’ law, the intensity through an analyser at angle θ is:

(a) I0 sinθ    (b) I0 cosθ    (c) I0 cos2θ    (d) I0 sin2θ

10. The Brewster angle for a medium of refractive index n is given by:

(a) sin iB = n    (b) cos iB = n    (c) tan iB = n    (d) tan iB = 1/n

Answer key: 1-(b), 2-(b), 3-(c), 4-(b), 5-(b), 6-(c), 7-(b), 8-(b), 9-(c), 10-(c).

Assertion–Reason Questions

For each question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: When light travels from air into glass, its frequency does not change.

Reason: The frequency of light is fixed by the source, while only the speed and wavelength change with the medium.

A-R 2. Assertion: Two independent bulbs cannot produce a sustained interference pattern.

Reason: Light from two independent bulbs has a constant phase difference at all times.

A-R 3. Assertion: In Young’s experiment, immersing the set-up in water reduces the fringe width.

Reason: The wavelength of light decreases in a denser medium, and fringe width is proportional to wavelength.

A-R 4. Assertion: Sound waves cannot be polarised.

Reason: Polarisation is shown only by transverse waves, and sound in air is a longitudinal wave.

A-R 5. Assertion: In interference and diffraction, the total energy of the light is conserved.

Reason: Energy is merely redistributed from dark regions to bright regions, with no net creation or destruction.

Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(A).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Thinking the frequency of light changes on refraction — it never does; only speed and wavelength change.
  • Mixing up the bright- and dark-fringe conditions — bright = nλ, dark = (n + ½)λ.
  • Confusing the single-slit minimum condition (a sinθ = nλ) with the double-slit maximum condition (d sinθ = nλ).
  • Forgetting that immersing the YDSE in a medium divides fringe width and angular width by n, because λ becomes λ/n.
  • Writing Malus’ law as cosθ instead of cos2θ, or forgetting the extra ½ factor when unpolarised light first passes through a polaroid.
  • Unit slips: convert nm, Å, mm and cm to metres before substituting in formulas.

How to score full marks in this chapter

Always state the formula, substitute with correct SI units, and box the final answer with its unit. For derivation questions (Snell’s law, YDSE fringe width) draw a neat labelled diagram and justify each geometric step. Remember the three “invariants” on refraction — frequency stays the same, speed and wavelength scale by 1/n. For polarisation numericals, apply the ½ factor for the first polaroid and Malus’ law for the next. Memorise the quick results: Brewster’s law tan iB = n and Fresnel distance zF = a2/λ.

Frequently Asked Questions

What is Class 12 Physics Chapter 10 Wave Optics about?

Chapter 10, Wave Optics, treats light as a wave. It covers wavefronts and Huygens’ principle, derivation of the laws of reflection and refraction, interference and Young’s double-slit experiment, single-slit diffraction, and polarisation (Malus’ law and Brewster’s angle) — phenomena that prove light is a transverse electromagnetic wave.

How many exercises are there in Class 12 Physics Chapter 10?

The NCERT 2026–27 textbook has 10 numbered Exercises (10.1 to 10.10). All of them are reproduced verbatim and solved step by step on this page, with every numerical worked out and verified to the correct units.

Why does fringe width decrease when the YDSE set-up is placed in water?

In water the wavelength becomes λ/n while the slit separation d and screen distance D are unchanged. Since fringe width β = λD/d, it reduces by the factor n (here 4/3), so the fringes move closer together.

Are these Class 12 Physics Chapter 10 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for the 2026–27 session.

← Chapter 9: Ray Optics All Class 12 Physics Chapter 11: Dual Nature of Radiation and Matter →
Scroll to Top