Wave Optics ISC Class 12: Notes, Derivations and Numericals

What Wave Optics is about

Weightage: Wave Optics is part of the unit Optics, which carries 18 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Wavefronts and Huygens' principle

A wavefront is a surface on which all points oscillate in the same phase. Rays are perpendicular to wavefronts. Shapes: a point source gives spherical wavefronts (in three dimensions, not circular), a line source cylindrical, and a source very far away (such as a star) plane wavefronts. Huygens' principle: (1) every point on a wavefront is a source of secondary wavelets, which spread out in the forward direction at the speed of the wave; (2) the envelope (common tangent surface) of these wavelets after a time t is the new wavefront. When light enters a denser medium its speed and wavelength fall in the same ratio while its frequency stays the same: $v=c/n$, ${\lambda }_m=\lambda /n$.

Worked example

Light of wavelength 600 nm in air enters glass of refractive index 1.5. Find its speed, wavelength and frequency in the glass.

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. ${\lambda }_g=\frac{600}{1.5}=400nm$
3. Frequency is unchanged: $f=\frac{c}{\lambda }=\frac{3\times 10^8}{600\times 10^{-9}}=5\times 10^{14}Hz$

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

Laws of reflection and refraction from Huygens' principle

Refraction: a plane wavefront AB meets the boundary at A at angle i. While the wavelet from B travels BC = $v_{1}t$ in medium 1, the wavelet from A travels AE = $v_{2}t$ in medium 2. The refracted wavefront CE is the tangent from C to the wavelet around A. In triangles ABC and AEC (common hypotenuse AC): $\sin i=\frac{BC}{AC}=\frac{v_{1}t}{AC}$ and $\sin r=\frac{AE}{AC}=\frac{v_{2}t}{AC}$. So $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}={}_1{n}_2$, a constant (Snell's law); if $v_2<v_1$ the ray bends towards the normal. Reflection: the same construction with the wavelet in the same medium gives AE = BC = vt, so triangles ABC and CEA are congruent and i = r.

Worked example

A plane wavefront in air strikes a glass surface (n = √3) at an angle of incidence of 60°. Find the angle of refraction and the speed of light in the glass.

Try it first, then show the solutionSolution

1. $\sin r=\frac{\sin 60^{\circ }}{\sqrt{3}}=\frac{0.866}{1.732}=0.5$, r = 30°
2. $v=\frac{c}{n}=\frac{3\times 10^8}{1.732}=1.73\times 10^{8}m{s}^{-1}$

Answer: r = 30°; $1.73\times 10^{8}m{s}^{-1}$

Interference: coherence, path and phase difference

When two waves meet, the resultant displacement is the sum of the individual displacements (superposition). Phase difference and path difference are related by $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$. Constructive interference (bright): $\Delta x=n\lambda$ ($\Delta \varphi =2n\pi$). Destructive (dark): $\Delta x=(n+\frac{1}{2})\lambda$ ($\Delta \varphi =(2n+1)\pi$). Coherent sources have the same frequency and a constant phase difference; two independent lamps are incoherent because their atoms emit randomly, so their phase difference changes millions of times a second and the pattern averages out. Conditions for sustained interference: coherent sources (derived from one source, as by two slits); same (or nearly the same) amplitude, for good contrast; sources close together and the screen far away, so fringes are wide enough to see; narrow sources; monochromatic light.

Worked example

Light of wavelength 500 nm from two coherent slits reaches point P with a path difference of 1.5 µm and point Q with 1.25 µm. Is each point bright or dark? What is the phase difference at Q?

Try it first, then show the solutionSolution

1. At P: $\frac{1.5\times 10^{-6}}{500\times 10^{-9}}=3$, a whole number: bright.
2. At Q: $\frac{1.25\times 10^{-6}}{500\times 10^{-9}}=2.5=(2+\frac{1}{2})$: dark.
3. $\Delta \varphi =2\pi \times 2.5=5\pi$ rad at Q.

Answer: P bright (3λ); Q dark (2.5λ), phase difference 5π

Young's double slit experiment and fringe width

Two narrow slits $S_1$ and $S_2$, a distance d apart, are lit by the same monochromatic source, so they are coherent. On a screen at distance D (D ≫ d), consider a point P at height y from the centre O. The path difference is $\Delta x=S_{2}P-S_{1}P=d\sin \theta$; for small θ, $\sin \theta \approx \tan \theta =\frac{y}{D}$, so $\Delta x=\frac{yd}{D}$. Bright fringes: $\frac{yd}{D}=n\lambda \Rightarrow y_n=\frac{nD\lambda }{d}$. Dark fringes: $y_n=(n+\frac{1}{2})\frac{D\lambda }{d}$. The fringe width (spacing between consecutive bright or dark fringes) is $\beta =\frac{\lambda D}{d}$, so fringes are equally spaced. β increases with λ and D and decreases with d; in a medium of index n it becomes β/n. The intensity graph shows equally bright maxima (ideally) of equal width. With white light the central fringe is white and the others are coloured.

Worked example

In a double slit experiment d = 0.5 mm, D = 1.0 m and λ = 600 nm. Find the fringe width, the distance of the third bright fringe from the centre and of the second dark fringe.

Try it first, then show the solutionSolution

1. $\beta =\frac{\lambda D}{d}=\frac{600\times 10^{-9}\times 1}{0.5\times 10^{-3}}=1.2\times 10^{-3}m=1.2mm$
2. Third bright: $y_3=3\beta =3.6mm$
3. Second dark: $y=(1+\frac{1}{2})\beta =1.8mm$ (counting the first dark fringe as n = 0)

Answer: 1.2 mm; 3.6 mm; 1.8 mm

Factors affecting the fringe pattern

From $\beta =\frac{\lambda D}{d}$: moving the screen away (larger D) widens the fringes; bringing the slits closer (smaller d) widens them; longer wavelength (red rather than blue) widens them. Immersing the whole apparatus in water reduces λ to λ/n, so β falls to β/n. Covering one slit destroys the interference; the screen then shows the single-slit diffraction pattern of the open slit (a broad central band), not uniform illumination. Widening the source slit too much washes the fringes out. Interference does not destroy energy: it redistributes it, with dark regions balanced by brighter ones.

Worked example

The apparatus of the previous example (β = 1.2 mm in air) is immersed in water of refractive index 4/3. Find the new fringe width.

Try it first, then show the solutionSolution

1. In water ${\lambda }^{\prime }=\lambda /n$, so ${\beta }^{\prime }=\beta /n$.
2. ${\beta }^{\prime }=\frac{1.2}{4/3}=0.9mm$

Answer: 0.9 mm

Single-slit Fraunhofer diffraction (qualitative)

Diffraction is the bending of light round the edges of an obstacle or aperture; it is noticeable only when the aperture width a is comparable to λ (which is why everyday diffraction of light is hard to see, while sound bends round doorways). In the Fraunhofer arrangement a parallel beam falls on a slit and the pattern is viewed with a lens on a screen in its focal plane (or on a distant screen). Dividing the slit into halves: at angle θ with $a\sin \theta =\lambda$, each point in the upper half has a partner in the lower half with path difference λ/2, so the light cancels: the first minimum. Generally minima at $a\sin {\theta }_n=n\lambda$ (n = 1, 2, 3, …) and secondary maxima roughly at $a\sin {\theta }_n=(n+\frac{1}{2})\lambda$. The central maximum runs between the first minima on either side: angular width $2{\theta }_1=\frac{2\lambda }{a}$, linear width $\frac{2D\lambda }{a}$ (twice that of a secondary maximum). Intensity falls sharply: the secondary maxima are much fainter than the central one. Narrowing the slit widens the pattern.

Worked example

Light of wavelength 600 nm falls on a slit 0.2 mm wide; the screen is 1 m away. Find the distance of the first minimum from the centre and the width of the central maximum.

Try it first, then show the solutionSolution

1. $y_1=\frac{D\lambda }{a}=\frac{1\times 600\times 10^{-9}}{0.2\times 10^{-3}}=3\times 10^{-3}m=3mm$
2. Central maximum width $=2y_1=6mm$ (angular width $2\lambda /a=6\times 10^{-3}rad$)

Answer: 3 mm; 6 mm

Interference compared with diffraction

Both arise from superposition, but: (1) interference is between waves from two coherent sources; diffraction is between secondary wavelets from different parts of the same wavefront. (2) Interference fringes (ideal YDSE) are of equal width; in diffraction the central maximum is twice as wide as the others. (3) Interference bright fringes are all of nearly equal intensity; diffraction maxima fall off rapidly in intensity. (4) Interference minima can be perfectly dark; diffraction minima are dark but the secondary maxima are faint, so contrast is poorer. In a real double-slit pattern the interference fringes are seen within the single-slit diffraction envelope.

Worked example

A student says: 'In single-slit diffraction all bright bands are as wide and as bright as the central band, just like YDSE fringes.' Correct the statement.

Try it first, then show the solutionSolution

1. The central maximum has width $2D\lambda /a$, twice that of each secondary maximum.
2. Intensity falls rapidly: the first secondary maximum is only a few per cent of the central intensity (about 4.7%), and later ones fainter still.

Answer: Wrong: the central band is twice as wide and far brighter than the secondary bands.

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

• Young's Double Slit Experiment and Fringe Width
• Huygens' Principle and Wavefronts
• Single Slit Diffraction and the Central Maximum

Wave Optics formula sheet

Speed and wavelength in a medium

$v=\frac{c}{n},{\lambda }_m=\frac{\lambda }{n}$

Frequency unchanged.

Snell's law from Huygens

$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}={}_1{n}_2$

Derived from the wavefront construction.

Phase and path difference

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

Bright: $\Delta x=n\lambda$; dark: $(n+\frac{1}{2})\lambda$.

YDSE path difference

$\Delta x=d\sin \theta \approx \frac{yd}{D}$

Small θ: $\sin \theta \approx \tan \theta =y/D$.

Fringe positions

$y_n=\frac{nD\lambda }{d}(bright);y_n=\left(n+\frac{1}{2}\right)\frac{D\lambda }{d}(dark)$

n = 0, 1, 2, …

Fringe width

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

In a medium: β/n. Angular fringe width λ/d.

Single-slit minima

$a\sin {\theta }_n=n\lambda ,n=1,2,3,\dots$

Secondary maxima near $(n+\frac{1}{2})\lambda$.

Central maximum

$2{\theta }_1=\frac{2\lambda }{a},width=\frac{2D\lambda }{a}$

Twice the width of a secondary maximum.

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.

• Point, line, far

  Point source → spherical; line source → cylindrical; far away → plane.

  Use it when: Wavefront-shape one-mark items.

• Bright is whole, dark is half

  Bright: path difference a whole number of wavelengths; dark: an odd half.

  Use it when: Deciding bright or dark from a path difference.

• Lambda-D over d

  β = λD/d: 'Long Distances widen; dense (close) slits widen too'.

  Use it when: Effect of changing λ, D or d.

• Opposite sides add

  Fringes on opposite sides of the centre: add their distances; same side: subtract.

  Use it when: Distance-between-fringes numericals.

• Single slit: minima at whole λ

  Single slit is the reverse of YDSE: $a\sin \theta =n\lambda$ gives DARK bands.

  Use it when: Not swapping the YDSE and single-slit conditions.

• Central band is double

  Diffraction's central maximum is twice as wide as the rest; YDSE fringes are all equal.

  Use it when: Comparing interference and diffraction.

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 5-mark question split into 1 + 1 + 1 + 2 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: Calling the wavefront of a point source 'circular'

  CISCE examiners' comments, 2024: for the wavefront of a point source, candidates wrote 'circular' or drew a circle.

  Wrong working

  Circular wavefronts.

  Right working

  Spherical wavefronts (light spreads in three dimensions).

  ✓Do this: Wavefronts are surfaces: spherical, cylindrical or plane.

  Usually costs: 1 mark.

• ✗Mistake: Proving reflection when refraction is asked, or a weak diagram

  CISCE examiners' comments, 2024 and 2025: for the Huygens proof of refraction, candidates proved the law of reflection, drew no diagram, missed arrows, did not mark i, r or the media, wrote sin i and sin r wrongly, or did not show that the ratio is constant.

  Wrong working

  Shows $\sin i=\sin r$, so i = r.

  Right working

  Wavelet in the second medium has radius $v_{2}t\ne v_{1}t$; $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$ = constant.

  ✓Do this: Read which law is asked; label media, wavefronts, i and r.

  Usually costs: 2-3 marks.

• ✗Mistake: Subtracting fringe distances on opposite sides

  CISCE examiners' comments, 2025: asked for the distance between the 5th bright fringe on one side and the 3rd on the other, candidates used the wrong formula, subtracted instead of adding, did not convert units, or ignored the second part on changing λ.

  Wrong working

  $(5-3)\beta =2\beta$

  Right working

  Opposite sides: $(5+3)\beta =8\beta$; same side: $(5-3)\beta$.

  ✓Do this: Sketch the centre and both fringes before calculating.

  Usually costs: 1-2 marks.

• ✗Mistake: Poor single-slit intensity graph

  CISCE examiners' comments, 2024: the x-axis was marked wrongly and the fall in intensity of secondary maxima away from the centre was not shown.

  Wrong working

  Equal peaks, like interference fringes.

  Right working

  Tall, broad central peak; much smaller secondary peaks falling outwards; minima at multiples of λ/a.

  ✓Do this: Label I and θ; mark at least two minima and two maxima on each side.

  Usually costs: 1 mark.

• ✗Mistake: Missing steps in the fringe-width derivation

  CISCE examiners' comments, 2024: diagrams were incomplete or unlabelled and the path-difference step or the fringe-width step was quoted wrongly.

  Wrong working

  Writes $\beta =\lambda D/d$ directly after $\Delta x=d\sin \theta$.

  Right working

  $\Delta x=yd/D$ (small angle), $y_n=nD\lambda /d$, then $\beta =y_{n+1}-y_n$.

  ✓Do this: Show each of the three steps.

  Usually costs: 1-2 marks.

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 wavefronts from a point source of light are:
   (a) plane
   (b) cylindrical
   (c) spherical
   (d) elliptical

   Hint

   Light spreads equally in all directions.

   Answer

   (c) spherical

2. Q2 · 1 mark · MCQBoard level

   In Young's experiment, the fringe width increases if:
   (a) the slits are moved further apart
   (b) the screen is moved further away
   (c) shorter-wavelength light is used
   (d) the apparatus is put in water

   Hint

   β = λD/d.

   Answer

   (b)

3. Q3 · 1 mark · MCQBoard level

   In single-slit diffraction, the central maximum is:
   (a) as wide as each secondary maximum
   (b) twice as wide as each secondary maximum
   (c) half as wide
   (d) dark

   Hint

   It runs from −λ/a to +λ/a.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   What are coherent sources?

   Hint

   Phase relation.

   Answer

   Sources that emit light of the same frequency with a constant phase difference.

5. Q5 · 2 marks · Short answerBoard level

   State Huygens' principle.

   Hint

   Two parts.

   Answer

   Every point on a wavefront acts as a source of secondary wavelets that travel forward at the wave speed; the new wavefront at a later time is the forward envelope (common tangent) of these wavelets.

6. Q6 · 2 marks · Short answerHarder board level

   Light of 600 nm passes through a slit 0.2 mm wide onto a screen 1 m away. Find the distance of the first minimum and the width of the central maximum.

   Hint

   $y_1=D\lambda /a$.

   Answer

   3 mm; 6 mm

7. Q7 · 2 marks · Short answerHarder board level

   A plane wavefront in air meets glass of refractive index √3 at 60° incidence. Find the angle of refraction and the speed of light in the glass.

   Hint

   Snell's law.

   Answer

   r = 30°; $v=1.73\times 10^{8}m{s}^{-1}$

8. Q8 · 2 marks · Short answerHOTS / competency

   What would you see in Young's experiment if monochromatic light were replaced by white light? Explain.

   Hint

   β depends on λ.

   Answer

   The central fringe is white (zero path difference for every colour). On either side a few coloured fringes appear, violet edge nearer the centre and red farther, because β ∝ λ; further out the colours overlap and the screen looks uniformly lit.

9. Q9 · 3 marks · Short answerHarder board level

   For a single slit of width a, find the angular width of the central maximum. Light of 600 nm passes through a 0.2 mm slit; find this angular width. What happens if the slit width is halved?

   Hint

   First minima at $\sin \theta =\pm \lambda /a$.

   Answer

   $2\lambda /a$; $2\times 600\times 10^{-9}/0.2\times 10^{-3}=6\times 10^{-3}rad$; halving a doubles it to $1.2\times 10^{-2}$ rad.

10. Q10 · 3 marks · Short answerHarder board level

    In a YDSE with d = 0.5 mm, D = 1 m and λ = 600 nm, find the distance between the fourth bright fringe on one side of the centre and the second bright fringe on the other side, and between the fourth and second bright fringes on the same side.

    Hint

    β first.

    Answer

    β = 1.2 mm; opposite sides 6β = 7.2 mm; same side 2β = 2.4 mm.

11. Q11 · 3 marks · Short answerHarder board level

    Using Huygens' principle, prove the law of reflection, with a labelled diagram.

    Hint

    Wavelet from A has radius vt in the same medium.

    Answer

    Incident wavefront AB; in time t, B travels BC = vt to the surface; wavelet from A has radius AE = vt. Reflected wavefront CE is tangent from C. Right triangles ABC and CEA share hypotenuse AC with BC = AE, so they are congruent and the angle of incidence equals the angle of reflection.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Describe Fraunhofer diffraction at a single slit, explaining the formation of the first minimum. [2]
    (b) Write the conditions for minima and secondary maxima, and sketch the intensity distribution. [2]
    (c) Give two differences between the patterns in single-slit diffraction and double-slit interference. [1]

    Hint

    Divide the slit into two halves.

    Answer

    (a) A parallel beam on a slit of width a is focused by a lens on a screen. At angle θ with $a\sin \theta =\lambda$, wavelets from each point of the upper half cancel those from the corresponding point of the lower half (path difference λ/2): first minimum. (b) Minima $a\sin \theta =n\lambda$; secondary maxima near $(n+\frac{1}{2})\lambda$; graph with a broad tall central peak and small decreasing side peaks. (c) Central maximum twice as wide (fringes equal in YDSE); intensity of maxima falls rapidly (nearly equal in YDSE).

13. Q13 · 2 marks · Short answerHarder board level

    A YDSE gives a fringe width of 1.2 mm with 600 nm light. Find the fringe width for 450 nm light and the ratio of the two fringe widths.

    Hint

    β ∝ λ.

    Answer

    0.9 mm; ratio 4 : 3.

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.Define a wavefront.Show answer

   The locus of all points vibrating in the same phase.

2. 2.State Huygens' principle.Show answer

   Each point on a wavefront is a source of secondary wavelets; their forward envelope gives the new wavefront.

3. 3.Conditions for constructive and destructive interference?Show answer

   Path difference nλ (bright) and $(n+\frac{1}{2})\lambda$ (dark).

4. 4.Fringe width in YDSE?Show answer

   $\beta =\lambda D/d$.

5. 5.Why are two independent lamps not coherent?Show answer

   Their phase difference changes randomly and rapidly.

6. 6.Condition for minima in single-slit diffraction?Show answer

   $a\sin \theta =n\lambda$, n = 1, 2, 3, …

7. 7.Angular width of the central maximum?Show answer

   $2\lambda /a$.

8. 8.What happens to frequency when light enters glass?Show answer

   It stays the same; speed and wavelength decrease.

Quick revision checklist

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

• ☐Can you name the wavefront for a point, a line and a distant source?
• ☐Can you prove Snell's law and the law of reflection with Huygens' construction?
• ☐Can you decide bright or dark from a path difference and find the phase difference?
• ☐Can you derive β = λD/d step by step?
• ☐Can you find the distance between fringes on the same side and on opposite sides?
• ☐Can you predict the effect of changing λ, D, d, the medium, or covering one slit?
• ☐Can you state the single-slit minima condition and the central-maximum width?
• ☐Can you draw the single-slit intensity graph correctly?

Wave Optics: frequently asked questions