JEE Main Physics · Chapter 19

Wave Optics

Huygens’ principle, Young’s double-slit experiment, diffraction, resolving power and polarisation.

NCERT Class 12JEE Main: 1 Q per shift 8 min read

1. Chapter at a Glance

Wave optics explains effects that ray optics cannot: interference, diffraction and polarisation. They show that light is a wave. The chapter covers Huygens' principle, coherent sources, Young's double-slit experiment (YDSE), single-slit diffraction, resolving power and polarisation (Malus's and Brewster's laws). JEE Main asks one question almost every year, usually on YDSE fringe width or intensity.

What JEE asks most
  • Fringe width and fringe positions in YDSE; effect of a medium or a thin sheet.
  • Resultant intensity and Imax/Imin for unequal sources.
  • Width of the central maximum in single-slit diffraction.
  • Malus's law and Brewster's angle.

2. Huygens' Principle

  • Every point on a wavefront acts as a source of secondary wavelets; the new wavefront is the forward envelope of these wavelets.
  • It explains the laws of reflection and refraction, and predicts that light slows down in a denser medium (v = c/n) — confirmed by experiment.
  • Point source → spherical wavefronts; distant source → plane wavefronts.

3. Interference and YDSE

  • Coherent sources have a constant phase difference (same frequency). Two independent bulbs are not coherent. YDSE gets coherence by dividing one wavefront.
  • Resultant intensity: I = I₁ + I₂ + 2√(I₁I₂) cosφ. For equal intensities, I = 4I₀ cos²(φ/2).
  • Imax/Imin = [(√I₁ + √I₂)/(√I₁ − √I₂)]² = [(a₁ + a₂)/(a₁ − a₂)]².
  • Path difference Δ = yd/D. Bright fringe: Δ = nλ; dark fringe: Δ = (2n − 1)λ/2.
  • Fringe width β = λD/d; in a medium of index μ, β′ = β/μ.
  • A thin sheet (thickness t, index μ) in front of one slit shifts the whole pattern by (μ − 1)tD/d towards that slit; number of fringes shifted = (μ − 1)t/λ.
  • With white light, the central fringe is white and the others are coloured.
S₁S₂ screen P y Dd path difference S₂P − S₁P = yd/D
YDSE: bright fringes where yd/D = nλ, giving equally spaced fringes of width β = λD/d.

4. Diffraction and Resolving Power

  • Diffraction is the bending of light around an obstacle or slit comparable in size to λ.
  • Single slit of width a: minima at a sinθ = nλ (n = 1, 2, …). The central maximum has angular width 2λ/a and linear width 2λD/a — twice the width of the secondary maxima.
  • Narrowing the slit widens the central maximum.
  • Resolving power: the smallest angle a telescope can resolve is θ ≈ 1.22λ/D (D = aperture diameter). Larger apertures and shorter wavelengths resolve finer detail. A microscope's resolving power is 2n sinθ/1.22λ.

5. Polarisation

  • Polarisation proves that light is a transverse wave (sound cannot be polarised).
  • Unpolarised light through a polariser: intensity halves.
  • Malus's law: I = I₀cos²θ for polarised light through an analyser at angle θ.
  • Brewster's law: at the polarising angle, tan θp = n; the reflected light is completely polarised and the reflected and refracted rays are perpendicular (θp + r = 90°).
  • Uses: Polaroid sunglasses reduce glare; 3D movies; stress analysis.

6. Formula Master Table

QuantityFormula
Resultant intensityI = I₁ + I₂ + 2√(I₁I₂) cosφ
Intensity ratioImax/Imin = ((a₁ + a₂)/(a₁ − a₂))²
Fringe widthβ = λD/d
nth bright / darky = nβ; y = (2n − 1)β/2
Sheet shift(μ − 1)tD/d
Single slit central max2λD/a
Resolution limitθ = 1.22λ/D
Malus / BrewsterI = I₀cos²θ; tanθp = n

7. Mind Map

Wave Optics
Huygens
  • Wavefronts
  • Reflection, refraction
Interference
  • Coherence
  • YDSE, β = λD/d
  • Thin sheet shift
Diffraction
  • Single slit
  • Resolving power
Polarisation
  • Malus's law
  • Brewster's angle

8. Tips and Tricks

Shortcuts
  • Intensity ratio 9 : 1 means amplitude ratio 3 : 1, so Imax/Imin = (4/2)² = 4.
  • When two wavelengths are used, bright fringes coincide where n₁λ₁ = n₂λ₂ — use the LCM.
  • Immersing YDSE in water shrinks β by the factor μ.
  • Two crossed polarisers block light; a third one at 45° between them lets I₀/8 through.
Mistakes students make
  • Forgetting to halve unpolarised light at the first polariser.
  • Using D for slit separation (d) or vice versa.
  • Thinking the central maximum in single-slit diffraction has the same width as the others.

9. Important One-liners

  • Interference needs coherent sources.
  • Energy is conserved in interference; it is only redistributed.
  • Polarisation shows that light waves are transverse.
  • At Brewster's angle the reflected and refracted rays are perpendicular.
  • Diffraction is noticeable when the obstacle size is comparable to the wavelength.
  • A bigger telescope aperture gives better resolution.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working.

Average level

N1Average

In YDSE, λ = 600 nm, slit separation d = 1 mm and screen distance D = 1 m. Find the fringe width.

Show answer & solution

Concept: β = λD/d.

  1. β = 600 × 10⁻⁹ × 1/10⁻³.
  2. = 6 × 10⁻⁴ m = 0.6 mm.

Answer: 0.6 mm

N2Average

The apparatus of N1 is immersed in water (μ = 4/3). Find the new fringe width.

Show answer & solution

Concept: λ in water = λ/μ, so β′ = β/μ.

  1. β′ = 0.6/(4/3) = 0.6 × 0.75.
  2. = 0.45 mm.

Answer: 0.45 mm

N3Average

Light of wavelength 500 nm passes through a single slit 0.1 mm wide onto a screen 2 m away. Find the width of the central maximum.

Show answer & solution

Concept: width = 2λD/a.

  1. Width = 2 × 5 × 10⁻⁷ × 2/10⁻⁴.
  2. = 0.02 m = 2 cm.

Answer: 2 cm

N4Average

Find Brewster's angle for glass of refractive index √3, and the angle of refraction at that incidence.

Show answer & solution

Concept: tanθp = n; θp + r = 90°.

  1. tanθp = √3 ⇒ θp = 60°.
  2. r = 90° − 60° = 30°.

Answer: 60°; 30°

N5Average

Unpolarised light of intensity I₀ passes through two polarisers whose axes are at 60°. Find the transmitted intensity.

Show answer & solution

Concept: first polariser halves; then Malus's law.

  1. After the first: I₀/2.
  2. After the second: (I₀/2)cos²60° = (I₀/2)(1/4).
  3. = I₀/8.

Answer: I₀/8

Advanced level

N6Advanced

Two coherent sources have intensities in the ratio 9 : 1. Find the ratio of maximum to minimum intensity in their interference pattern.

Show answer & solution

Concept: amplitude ∝ √intensity.

  1. a₁ : a₂ = 3 : 1.
  2. Imax ∝ (3 + 1)² = 16; Imin ∝ (3 − 1)² = 4.
  3. Ratio = 4 : 1.

Answer: 4 : 1

N7Advanced

In YDSE with λ = 600 nm, a thin glass sheet (μ = 1.5) placed over one slit shifts the central fringe by 5 fringes. Find the sheet's thickness.

Show answer & solution

Concept: extra path (μ − 1)t = number of fringes × λ.

  1. (1.5 − 1)t = 5 × 600 × 10⁻⁹.
  2. t = 3 × 10⁻⁶/0.5 = 6 × 10⁻⁶ m.
  3. = 6 μm.

Answer: 6 μm

N8Advanced

For the set-up of N1 (β = 0.6 mm), find the distance from the centre of (a) the 3rd bright fringe, (b) the 3rd dark fringe, and (c) the separation between them.

Show answer & solution

Concept: bright y = nβ; dark y = (2n − 1)β/2.

  1. (a) y₃(bright) = 3 × 0.6 = 1.8 mm.
  2. (b) y₃(dark) = 5 × 0.6/2 = 1.5 mm.
  3. (c) Separation = 0.3 mm (= β/2).

Answer: 1.8 mm; 1.5 mm; 0.3 mm

N9Advanced

A YDSE (d = 1 mm, D = 1 m) is lit by 600 nm and 400 nm together. Find the nearest point to the centre where bright fringes of both wavelengths coincide.

Show answer & solution

Concept: coincidence when n₁λ₁ = n₂λ₂.

  1. 600n₁ = 400n₂ ⇒ n₁/n₂ = 2/3; smallest: n₁ = 2, n₂ = 3.
  2. y = n₁λ₁D/d = 2 × 600 × 10⁻⁹ × 1/10⁻³.
  3. y = 1.2 × 10⁻³ m = 1.2 mm.

Answer: 1.2 mm from the centre

N10Advanced

A telescope has an objective of diameter 10 cm. Find the smallest angular separation it can resolve for light of 500 nm. Would doubling the diameter help?

Show answer & solution

Concept: Rayleigh criterion θ = 1.22λ/D.

  1. θ = 1.22 × 5 × 10⁻⁷/0.1.
  2. = 6.1 × 10⁻⁶ rad (about 1.3 arc-seconds).
  3. Doubling D halves θ — the telescope resolves twice as finely.

Answer: about 6.1 × 10⁻⁶ rad; yes, it halves the limit

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

Two sources are coherent if they have:

  1. Same amplitude only
  2. Constant phase difference
  3. Different frequencies
  4. Same intensity only
Show solution

Coherence requires the same frequency and a constant phase difference.

Answer: (B)

Q2Easy

Which phenomenon proves that light is a transverse wave?

  1. Interference
  2. Diffraction
  3. Polarisation
  4. Reflection
Show solution

Only transverse waves can be polarised.

Answer: (C)

Q3Medium

In YDSE, the slit separation is doubled and the screen distance is halved. The fringe width becomes:

  1. Unchanged
  2. Half
  3. One-fourth
  4. Four times
Show solution

β ∝ D/d = (1/2)/2 = 1/4.

Answer: (C)

Q4Hard

In interference, the energy at dark fringes:

  1. Is destroyed
  2. Is converted to heat
  3. Is redistributed to bright fringes
  4. Goes back to the source
Show solution

Energy is conserved; average intensity over the pattern equals I₁ + I₂. Bright fringes get the missing energy.

Answer: (C)

Q5Numerical

Find Brewster's angle (in degrees) for a medium of refractive index √3.

Show solution

tanθp = √3 ⇒ 60°.

Answer: 60

Q6Numerical

Two coherent waves each of intensity I₀ interfere with phase difference 60°. Find the resultant intensity in units of I₀.

Show solution

I = 4I₀cos²(30°) = 4I₀ × 3/4 = 3I₀.

Answer: 3

12. Quick Revision Summary

Remember before the exam
  • Huygens: secondary wavelets; v = c/n in a medium.
  • I = I₁ + I₂ + 2√(I₁I₂)cosφ; Imax/Imin = ((a₁ + a₂)/(a₁ − a₂))².
  • β = λD/d; β/μ in a medium; sheet shift (μ − 1)tD/d.
  • Single slit: central width 2λD/a; resolution 1.22λ/D.
  • Malus I₀cos²θ; Brewster tanθp = n.
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