Wave Optics Class 12: Notes, Proofs and Solved Numericals

What Wave Optics is about

Weightage: This chapter is part of Units V and VI, Electromagnetic Waves; Optics, which carry 18 of the 70 theory marks together with Electromagnetic Waves and Ray Optics and Optical Instruments.Source: CBSE Physics syllabus 2026-27, Units V and VI Electromagnetic Waves; Optics

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.1 marks per paper. See the questions

Key concepts

Wavefronts and Huygens principle

NCERT section 10.2

A wavefront is a surface of constant phase: all points on it are vibrating in step. Rays are perpendicular to wavefronts and show the direction in which energy travels.

• A point source gives spherical wavefronts.
• Far from the source, a small part of a spherical wavefront is nearly flat: a plane wavefront (for example, light from the sun or a distant star).
• A line source (such as a slit) gives cylindrical wavefronts.
• A convex lens turns a plane wavefront into a converging spherical one; a prism turns it but keeps it plane.

Huygens principle: every point on a wavefront acts as a source of secondary wavelets, which spread out in the forward direction at the speed of the wave. The new position of the wavefront after time t is the common tangent (envelope) to all these wavelets. Huygens assumed the backward wave to be absent.

Worked example

What is the shape of the wavefront (a) from a point source very close to you, (b) of light from a distant star, (c) emerging from a convex lens when a point source is placed at its focus? [1 mark]

Try it first, then show the solutionSolution

1. (a) Close to a point source: spherical (diverging).
2. (b) A distant star: plane, because a small part of a very large sphere is nearly flat.
3. (c) Source at the focus: the lens makes the rays parallel, so the wavefront is plane.

Answer: (a) spherical (b) plane (c) plane

Refraction of a plane wave: proof of Snell's law

NCERT section 10.3.1 and 10.3.2

A plane wavefront AB meets the boundary between medium 1 (speed $v_1$) and medium 2 (speed $v_2$) at angle i. While the end B travels to C on the boundary in time τ, covering $BC=v_1\tau$, the secondary wavelet from A spreads into medium 2 to a radius $AE=v_2\tau$. The refracted wavefront is the tangent CE from C to this wavelet, making angle r with the boundary.

From the right-angled triangles ABC and AEC: $\sin i=\frac{BC}{AC}=\frac{v_1\tau }{AC}$ and $\sin r=\frac{AE}{AC}=\frac{v_2\tau }{AC}$. So

$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=n_{21}$

which is Snell's law, with the extra result that $n=\frac{c}{v}$: light travels slower in the denser medium (Newton's corpuscular theory predicted the opposite).

The frequency stays the same (it is set by the source), so the wavelength changes in proportion to speed: ${\lambda }_2=\frac{{\lambda }_1{v}_2}{v_1}$, or in a medium of index n, ${\lambda }^{\prime }=\frac{\lambda }{n}$.

Going into a rarer medium ($v_2>v_1$), r > i; at the critical angle r = 90°, and beyond it total internal reflection occurs.

Worked example

Light of wavelength 600 nm in air enters glass of refractive index 1.5. Find its speed, wavelength and frequency in glass. [2 marks]

Try it first, then show the solutionSolution

1. $v=\frac{c}{n}=\frac{3\times 10^8}{1.5}=2\times 10^{8}m{s}^{-1}$
2. ${\lambda }^{\prime }=\frac{\lambda }{n}=\frac{600}{1.5}=400nm$
3. Frequency is unchanged: $\nu =\frac{c}{\lambda }=\frac{3\times 10^8}{600\times 10^{-9}}=5\times 10^{14}Hz$

Answer: $2\times 10^{8}m{s}^{-1}$; 400 nm; $5\times 10^{14}Hz$

Reflection of a plane wave: proof of the law of reflection

NCERT section 10.3.3

A plane wavefront AB strikes a reflecting surface MN at angle i. While the end B travels to C in time τ, $BC=v\tau$, the secondary wavelet from A travels back into the same medium to a radius $AE=v\tau$ (same speed). The reflected wavefront is the tangent CE from C to this wavelet.

Triangles ABC and AEC are right-angled ($\angle B=\angle E=90^{\circ }$), share the hypotenuse AC, and have $BC=AE$. So they are congruent, and the angle the incident wavefront makes with the surface equals the angle the reflected wavefront makes with it: $i=r$. This is the law of reflection. (The incident ray, reflected ray and normal also lie in one plane.)

Worked example

A plane wavefront falls on a plane mirror. Using Huygens' construction, explain why the reflected wavefront travels at the same speed and why the angles of incidence and reflection are equal. [2 marks]

Try it first, then show the solutionSolution

1. The secondary wavelets from points on the mirror travel back into the same medium, so they move at the same speed v as the incident wave.
2. In the same time τ, the incident wavefront's free end moves $BC=v\tau$ and the wavelet from A spreads to $AE=v\tau$.
3. Right triangles ABC and AEC have a common hypotenuse AC and BC = AE, so they are congruent, giving i = r.

Answer: Same medium means same speed; congruent triangles give i = r.

Coherent sources and interference (superposition)

NCERT section 10.4

When two waves overlap, the resultant displacement is the sum of the individual displacements (superposition principle). For two waves of equal amplitude and phase difference φ, the resultant intensity is

$I=4I_0{\cos }^2\frac{\varphi }{2}$

where $I_0$ is the intensity of each wave alone.

• Constructive interference (bright): phase difference $\varphi =0,2\pi ,4\pi ,\dots$, i.e. path difference $n\lambda$; $I=4I_0$.
• Destructive interference (dark): $\varphi =\pi ,3\pi ,\dots$, i.e. path difference $(n+\frac{1}{2})\lambda$; $I=0$.

Coherent sources keep a constant phase difference with each other, so the pattern is steady (sustained interference). Two independent sources (two bulbs) are incoherent: their phase difference changes randomly many millions of times a second, the ${\cos }^2$ term averages to $\frac{1}{2}$, and the intensity is just $2I_0$ everywhere, with no fringes. Coherent sources are made by dividing one source into two, as in Young's experiment. Energy is not destroyed in interference; it is redistributed from dark regions to bright ones.

For waves of unequal intensities $I_1$ and $I_2$: $I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$ and $I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$.

Worked example

Two coherent waves have amplitudes in the ratio 3 : 1. Find the ratio of maximum to minimum intensity in their interference pattern. [2 marks]

Try it first, then show the solutionSolution

1. Intensity ∝ amplitude², and $I_{max}\propto (a_1+a_2{)}^2$, $I_{min}\propto (a_1-a_2{)}^2$
2. $\frac{I_{max}}{I_{min}}=\frac{(3+1{)}^2}{(3-1{)}^2}=\frac{16}{4}$
3. $=4:1$

Answer: 4 : 1

Young's double-slit experiment and fringe width

NCERT section 10.5

Light from a narrow slit S falls on two close, parallel slits $S_1$ and $S_2$ (separation d). These act as two coherent sources. On a screen at distance D (with $D\gg d$), bright and dark fringes appear, parallel to the slits and equally spaced.

At a point P on the screen, the path difference is $S_{2}P-S_{1}P$. P is bright when this equals $n\lambda$ and dark when it equals $(n+\frac{1}{2})\lambda$. The centre O (path difference zero) is bright.

The fringe width (distance between two successive bright or dark fringes) is

$\beta =\frac{\lambda D}{d}$

(final expression only; its derivation is not in the 2026-27 syllabus). Consequences:

• β increases with λ (red fringes are wider than blue) and with D; it decreases as d increases.
• In a medium of refractive index n, λ becomes $\frac{\lambda }{n}$, so ${\beta }^{\prime }=\frac{\beta }{n}$.
• With white light, the central fringe is white and the others are coloured, with only a few visible.
• If one slit is covered, the interference fringes disappear (only the single-slit pattern remains).
• The angular fringe width is $\frac{\lambda }{d}$.

Worked example

In a Young's double-slit experiment, light of wavelength 600 nm is used, the slits are 0.3 mm apart and the screen is 1.5 m away. Find (a) the fringe width, (b) the distance of the third bright fringe from the centre, (c) the fringe width if the apparatus is immersed in water of refractive index $\frac{4}{3}$. [3 marks]

Try it first, then show the solutionSolution

1. (a) $\beta =\frac{\lambda D}{d}=\frac{600\times 10^{-9}\times 1.5}{0.3\times 10^{-3}}=3\times 10^{-3}m=3mm$
2. (b) Third bright fringe: $3\beta =9mm$ from the centre
3. (c) ${\beta }^{\prime }=\frac{\beta }{n}=\frac{3}{4/3}=2.25mm$

Answer: (a) 3 mm (b) 9 mm (c) 2.25 mm

Diffraction at a single slit (qualitative)

NCERT section 10.6 and 10.6.1

When light passes through a single slit of width a comparable to its wavelength, it spreads out beyond the geometrical shadow: diffraction. On a distant screen you see a broad, bright central maximum with weaker, narrower maxima on either side, separated by dark minima.

Qualitative explanation: divide the slit into many small parts, each a source of secondary wavelets (Huygens). Straight ahead (θ = 0), all parts are in phase, so the intensity is maximum. At the angle where the path difference between the two edges of the slit is λ, the slit can be divided into two halves whose wavelets cancel in pairs: the first minimum, at $\theta \approx \frac{\lambda }{a}$. In general, minima occur at $\theta \approx \frac{n\lambda }{a}$.

Width of the central maximum: it extends from $-\frac{\lambda }{a}$ to $+\frac{\lambda }{a}$, so its angular width is $\frac{2\lambda }{a}$ and its linear width on a screen at distance D is $\frac{2\lambda D}{a}$. A narrower slit gives a wider central maximum; a longer wavelength also widens it. (The treatment is qualitative only for 2026-27.)

Interference versus diffraction: interference is the superposition of waves from two (or a few) separate coherent sources and gives equally spaced fringes of equal intensity; diffraction is the superposition of wavelets from different parts of one wavefront and gives a central maximum twice as wide as the others, with intensity falling off rapidly.

Worked example

Light of wavelength 500 nm falls on a slit 0.1 mm wide. Estimate the angular width and the linear width of the central maximum on a screen 1 m away. What happens if the slit width is halved? [2 marks]

Try it first, then show the solutionSolution

1. First minimum at $\theta \approx \frac{\lambda }{a}=\frac{500\times 10^{-9}}{10^{-4}}=5\times 10^{-3}rad$; angular width $=2\theta =0.01rad$
2. Linear width $=\frac{2\lambda D}{a}=0.01\times 1=0.01m=1cm$
3. Halving the slit width doubles the width of the central maximum, to 2 cm.

Answer: 0.01 rad; 1 cm; it doubles to 2 cm

Physics in real life: Wave Optics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Huygens Principle and Wavefronts
• Young's Double Slit Experiment
• Single Slit Diffraction

Wave Optics formula sheet

Snell's law from Huygens

$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=n_{21}$

Wavelength and speed in a medium

$v=\frac{c}{n},{\lambda }^{\prime }=\frac{\lambda }{n},{\nu }^{\prime }=\nu$

Phase and path difference

$\varphi =\frac{2\pi }{\lambda }\Delta x$

Resultant intensity (equal waves)

$I=4I_0{\cos }^2\frac{\varphi }{2}$

Unequal waves

$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2,I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

Conditions for fringes

$\Delta x=n\lambda \text{(bright)},\Delta x=\left(n+\frac{1}{2}\right)\lambda \text{(dark)}$

Fringe width (final expression only)

$\beta =\frac{\lambda D}{d}$

In a medium: $\beta /n$. Angular fringe width $\lambda /d$.

Position of the nth bright fringe

$y_n=n\beta =\frac{n\lambda D}{d}$

Single slit: minima (qualitative)

$\theta \approx \frac{n\lambda }{a}$

Width of the central maximum

$\text{angular: }\frac{2\lambda }{a},\text{linear: }\frac{2\lambda D}{a}$

Wave Optics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Wave Optics

The question patterns to expect, and the method that works for each.

Tricks to remember Wave Optics

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Lambda D by d

  'Lady Diana divided': β = λD/d. Anything on top (λ, D) widens the fringes; d below narrows them.

  Use it when: Fringe width and its changes.

• Frequency is the source's signature

  Entering a new medium, light keeps its frequency (its signature) but changes speed and wavelength by the factor n.

  Use it when: Wavelength in a medium questions.

• Bright = whole λ, dark = half λ

  Path difference a whole number of wavelengths: bright. An odd number of half-wavelengths: dark.

  Use it when: Fringe conditions.

• Narrow slit, wide centre

  Central maximum width = 2λD/a: halve the slit, double the spread.

  Use it when: Diffraction MCQs.

• 4I₀ cos²(φ/2)

  Two equal waves: from 4I₀ when in step to 0 when opposite; the average is 2I₀ (energy is conserved, just moved around).

  Use it when: Intensity at a point.

Wave Optics: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Wave Optics

The slips examiners see most often, and what to write instead.

• ✗Mistake: Deriving the fringe width at length

  Older syllabi required it.

  ✓Do this: For 2026-27 write the final expression $\beta =\lambda D/d$ and use it; spend time on the Huygens proofs instead.

  Usually costs: Time.

• ✗Mistake: In the Huygens refraction proof, using $v_1\tau$ for the wavelet in medium 2

  Not noticing which part is in which medium.

  Wrong working

  AE = $v_1\tau$

  Right working

  BC = $v_1\tau$ (still in medium 1); AE = $v_2\tau$ (wavelet in medium 2)

  ✓Do this: Label the two media and speeds on the diagram before writing the triangles.

  Usually costs: 1-2 marks.

• ✗Mistake: Saying frequency changes in a denser medium

  Confusing it with wavelength.

  ✓Do this: Frequency stays the same; $v$ and λ fall by the factor n.

  Usually costs: 1 mark.

• ✗Mistake: Taking the nth dark fringe at $n\beta$

  Mixing up bright and dark positions.

  Wrong working

  3rd dark fringe at $3\beta$

  Right working

  3rd dark fringe at $(3-\frac{1}{2})\beta =2.5\beta$

  ✓Do this: Bright: $n\beta$; dark: $(n-\frac{1}{2})\beta$ counted from the centre.

  Usually costs: 1 mark.

• ✗Mistake: Writing that energy is destroyed at dark fringes

  Misreading destructive interference.

  ✓Do this: Energy is redistributed: dark regions get zero, bright regions get 4I₀, average 2I₀.

  Usually costs: 1 mark in reasoning items.

• ✗Mistake: Answering polarisation questions from older papers

  Old question banks include them.

  ✓Do this: Polarisation is not in the 2026-27 syllabus; do not spend revision time on Brewster's law or Malus's law.

  Usually costs: Time.

NCERT exercise guide: Wave Optics

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Wave Optics: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.1 marks per paper; 26 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Wave Optics

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Wave Optics question, by year

Practice questions on Wave Optics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The wavefront of light from a distant star reaching the earth is:
   (a) spherical
   (b) cylindrical
   (c) plane
   (d) elliptical

   Hint

   A tiny part of a huge sphere.

   Answer

   (c) plane

2. Q2 · 1 mark · MCQBoard level

   In Young's experiment, if blue light is replaced by red light, the fringe width:
   (a) decreases
   (b) increases
   (c) stays the same
   (d) becomes zero

   Hint

   β ∝ λ.

   Answer

   (b) increases

3. Q3 · 1 mark · Very short answerBoard level

   Define coherent sources.

   Hint

   Phase difference.

   Answer

   Two sources of light that maintain a constant phase difference between them (and have the same frequency).

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): Two independent sodium lamps placed side by side do not produce an interference pattern.
   Reason (R): The phase difference between light from two independent sources changes rapidly and randomly.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   Coherence.

   Answer

   (a)

5. Q5 · 2 marks · Very short answerHarder board level

   Two coherent sources have intensities $I_0$ and $4I_0$. Find the maximum and minimum intensity in their interference pattern.

   Hint

   Use square roots.

   Answer

   $I_{max}=(\sqrt{I_0}+2\sqrt{I_0}{)}^2=9I_0$; $I_{min}=(2\sqrt{I_0}-\sqrt{I_0}{)}^2=I_0$.

6. Q6 · 2 marks · Very short answerHarder board level

   In Young's experiment the fringe width is 2 mm with light of 500 nm. What is it with 600 nm light, everything else unchanged? What would it be with 500 nm light if the slit separation is doubled?

   Hint

   β ∝ λ/d.

   Answer

   2.4 mm with 600 nm; 1 mm when d is doubled.

7. Q7 · 2 marks · Very short answerHarder board level

   Use Huygens principle to explain why a plane wave entering a denser medium bends towards the normal.

   Hint

   Wavelets in the denser medium travel less in the same time.

   Answer

   The part of the wavefront that enters the denser medium first sends out wavelets that travel more slowly, so in the same time they cover less distance than the part still in the rarer medium. The new wavefront is therefore turned, and the ray (perpendicular to it) bends towards the normal: r < i.

8. Q8 · 5 marks · Long answerHarder board level

   (a) State Huygens principle and use it to prove the law of refraction. (b) Light travels from glass (n = 1.5) into water (n = 1.33). Is it bent towards or away from the normal? Find the critical angle for this pair.

   Hint

   Going to a rarer medium.

   Answer

   (a) As in the standard construction: $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$. (b) Water is optically rarer than glass, so the ray bends away from the normal. $\sin i_c=\frac{1.33}{1.5}\approx 0.887$, $i_c\approx 62.5^{\circ }$.

9. Q9 · 2 marks · Very short answerHOTS / competency

   In Young's experiment the central maximum has intensity $I_m$. At a point on the screen the path difference is $\frac{\lambda }{6}$. Find the intensity there. Where is the intensity half the maximum?

   Hint

   $I=I_m{\cos }^2(\varphi /2)$.

   Answer

   $\varphi =\frac{\pi }{3}$, $I=I_m{\cos }^{2}30^{\circ }=\frac{3I_m}{4}$. Half intensity when ${\cos }^2(\varphi /2)=\frac{1}{2}$, $\varphi =\frac{\pi }{2}$, i.e. path difference $\frac{\lambda }{4}$ (a quarter of a fringe from a bright fringe).

10. Q10 · 2 marks · Very short answerHOTS / competency

    A single slit of width a is illuminated by light of wavelength λ. For what slit width will the first minimum be at an angle of 30°? Comment on the result.

    Hint

    $a\sin \theta =\lambda$ for the first minimum.

    Answer

    $a\sin 30^{\circ }=\lambda$, so $a=2\lambda$. The slit is only two wavelengths wide, so the light spreads out a lot: diffraction is strong when the opening is comparable to the wavelength.

11. Q11 · 3 marks · Short answerHarder board level

    Give three differences between the interference pattern of Young's experiment and the diffraction pattern of a single slit.

    Hint

    Sources, widths, intensities.

    Answer

    (1) Interference: superposition of waves from two coherent sources; diffraction: superposition of wavelets from different parts of a single wavefront. (2) Interference fringes are of equal width; in diffraction the central maximum is twice as wide as the others. (3) Interference bright fringes are of (nearly) equal intensity; diffraction maxima get rapidly fainter away from the centre.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State Huygens principle.Show answer

   Every point on a wavefront is a source of secondary wavelets spreading forward at the wave speed; the new wavefront is the forward envelope of these wavelets.

2. 2.What is a wavefront? Give the shape for a distant source.Show answer

   A surface of constant phase; plane for a distant source.

3. 3.What happens to frequency, speed and wavelength when light enters glass?Show answer

   Frequency unchanged; speed and wavelength reduced by the factor n.

4. 4.What are coherent sources?Show answer

   Sources that maintain a constant phase difference (usually derived from one source).

5. 5.Write the fringe width in Young's experiment.Show answer

   $\beta =\lambda D/d$.

6. 6.What is the intensity at a point where two equal coherent waves have a phase difference of $\frac{\pi }{2}$?Show answer

   $4I_0{\cos }^2\frac{\pi }{4}=2I_0$.

7. 7.Width of the central maximum in single-slit diffraction?Show answer

   Angular $2\lambda /a$; linear $2\lambda D/a$.

8. 8.Give two differences between interference and diffraction.Show answer

   Interference: two sources, equally spaced fringes of equal intensity. Diffraction: one wavefront, central maximum twice as wide, intensity falling for outer maxima.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Huygens principle and draw wavefronts for a point source and a distant source?
• ☐Can you prove Snell's law with Huygens construction?
• ☐Can you prove the law of reflection with Huygens construction?
• ☐Can you find speed, wavelength and frequency of light in a medium?
• ☐Can you explain why two bulbs do not produce interference fringes?
• ☐Can you find the intensity at a point from the path difference?
• ☐Can you find $I_{max}/I_{min}$ from amplitude or slit-width ratios?
• ☐Can you use β = λD/d and say how β changes with λ, D, d and the medium?
• ☐Can you find where bright fringes of two wavelengths first coincide?
• ☐Can you describe the single-slit pattern and the width of its central maximum?

Wave Optics: frequently asked questions