Wave optics for NEET

Wave optics is a compact chapter, and in NEET it is mostly about three results: the fringe width in Young's double-slit experiment, the width of the central maximum in single-slit diffraction, and the intensity after a polariser. Add the intensity formula for interference and Brewster's angle, and you have covered almost every question type.

The chapter rewards ratio thinking. Most questions change one thing (the slit gap, the wavelength, the medium) and ask what happens to the pattern.

Wavefronts and Huygens' principle

A wavefront is a surface on which every point is in the same phase. A point source gives spherical wavefronts; a line source gives cylindrical ones; far from any source, they are effectively plane. Rays are perpendicular to wavefronts.

Huygens' principle: every point on a wavefront acts as a source of secondary wavelets moving forward at the wave's speed. The new wavefront is the forward envelope of these wavelets.

Using this construction, NCERT derives the laws of reflection and refraction. Refraction gives n₁/n₂ = v₂/v₁, which is Snell's law. It also shows that when light enters a denser medium:

• the frequency stays the same,
• the speed falls to c/n,
• the wavelength falls to λ/n.

Coherence and interference

Two sources are coherent if they keep a constant phase difference. Only coherent sources give a steady (sustained) interference pattern. Two separate bulbs are not coherent, because each emits light in random, rapidly changing phases. So YDSE splits light from one source into two.

When two waves of intensities I₁ and I₂ meet with phase difference φ, the resultant intensity is

I = I₁ + I₂ + 2√(I₁I₂) cos φ

• Maximum: (√I₁ + √I₂)². Minimum: (√I₁ − √I₂)².
• For equal intensities I₀: I = 4I₀ cos²(φ/2). The maxima are 4I₀ and the minima are zero.
• Phase and path difference are linked by φ = (2π/λ) × path difference.

Constructive interference needs a path difference of nλ; destructive needs (n + ½)λ. Energy is not destroyed at the dark fringes; it is redistributed to the bright ones.

Young's double-slit experiment

Slits a distance d apart are lit by coherent light, and the pattern is seen on a screen at distance D (with D much larger than d). At a point y from the centre, the path difference is yd/D.

• Bright fringes at y = nλD/d.
• Dark fringes at y = (n − ½)λD/d, for n = 1, 2, 3…
• Fringe width β = λD/d, the same for bright and dark fringes.
• Angular fringe width = λ/d.

Diffraction at a single slit

Light passing through a single slit of width a spreads out and forms a broad central bright band with weaker bands on either side.

• Minima at a sin θ = nλ (n = 1, 2, 3…).
• The central maximum spans from the first minimum on one side to the first on the other: angular width 2λ/a, linear width 2λD/a on a screen at distance D.
• The secondary maxima lie roughly midway between minima and get fainter as you move out.

Diffraction is noticeable only when the slit width is comparable to the wavelength. That is why sound, with wavelengths of metres, bends around doors, but light casts sharp shadows.

Polarisation

Polarisation shows that light is a transverse wave; longitudinal waves such as sound cannot be polarised.

• Ordinary (unpolarised) light passing through a polaroid emerges plane-polarised, with half the intensity: I₀/2.
• Malus's law: when plane-polarised light of intensity I₀ passes through an analyser at angle θ to its plane, I = I₀ cos²θ. Crossed polaroids (θ = 90°) give zero.
• Brewster's law: light reflected from a transparent surface is completely plane-polarised when tan θ_B = n. At this angle, the reflected and refracted rays are at 90° to each other, so θ_B + r = 90°.

Uses of polaroids: sunglasses that cut glare, camera filters, LCD screens and 3D film viewing.

Worked numericals

Example 1: fringe width in air and water

λ = 600 nm, D = 1 m, d = 0.5 mm.

• β = (6 × 10⁻⁷ × 1)/(5 × 10⁻⁴) = 1.2 × 10⁻³ m = 1.2 mm.
• In water (n = 4/3): β = 1.2 × 3/4 = 0.9 mm.

Example 2: intensity ratio

Two slits give intensities in the ratio 9 : 1. Find I_max : I_min.

• Amplitude ratio = √9 : √1 = 3 : 1.
• I_max : I_min = (3 + 1)² : (3 − 1)² = 16 : 4 = 4 : 1.

Example 3: width of the central maximum

Light of 500 nm falls on a slit 0.2 mm wide, with the screen 2 m away.

• Width = 2λD/a = (2 × 5 × 10⁻⁷ × 2)/(2 × 10⁻⁴) = (2 × 10⁻⁶)/(2 × 10⁻⁴) = 10⁻² m = 1 cm.

Example 4: three polaroids

Unpolarised light of intensity I₀ passes through three polaroids. The first and third are crossed; the middle one is at 45° to both.

• After the first: I₀/2.
• After the second: (I₀/2) × cos² 45° = I₀/4.
• After the third: (I₀/4) × cos² 45° = I₀/8.

Without the middle polaroid, the crossed pair would transmit nothing. Adding a polaroid lets light through.

Example 5: shift due to a thin sheet

A thin sheet of index 1.5 placed over one slit shifts the central fringe by 5 fringe widths, with λ = 600 nm. Find its thickness.

• Extra path = (n − 1)t = 5λ.
• t = 5 × 600 nm / 0.5 = 6,000 nm = 6 μm.

Practice MCQs

1. In YDSE, the slit separation is doubled. The fringe width: (a) doubles (b) halves (c) is unchanged (d) becomes four times
2. A polariser and analyser are at 90°. The transmitted intensity is: (a) I₀ (b) I₀/2 (c) I₀/4 (d) zero
3. Unpolarised light of intensity I₀ passes through one polaroid. The transmitted intensity is: (a) I₀ (b) I₀/2 (c) I₀/4 (d) zero
4. In YDSE with β = 0.2 mm, the distance between the 3rd bright fringe and the 5th dark fringe on the same side of the centre is: (a) 0.1 mm (b) 0.2 mm (c) 0.3 mm (d) 0.4 mm
5. Two coherent sources each of intensity I₀ interfere. The maximum intensity is: (a) I₀ (b) 2I₀ (c) 4I₀ (d) 8I₀
6. Brewster's angle for glass of refractive index √3 is: (a) 30° (b) 45° (c) 60° (d) 90°
7. In single-slit diffraction, the slit width is halved. The width of the central maximum: (a) halves (b) doubles (c) is unchanged (d) becomes four times
8. Light of wavelength 600 nm in air enters glass of index 1.5. Its wavelength in glass is: (a) 900 nm (b) 600 nm (c) 400 nm (d) 300 nm

Answers

1. (b) β ∝ 1/d.
2. (d) cos² 90° = 0.
3. (b) A polaroid passes half of unpolarised light.
4. (c) 5th dark is at 4.5β, 3rd bright at 3β; the gap is 1.5β = 0.3 mm.
5. (c) (√I₀ + √I₀)² = 4I₀.
6. (c) tan θ_B = √3.
7. (b) Width ∝ 1/a.
8. (c) λ/n = 600/1.5 = 400 nm; the frequency is unchanged.

What to do next

• Rebuild the YDSE change table from β = λD/d, one row at a time.
• Solve five questions on fringe positions, counting bright and dark fringes from the centre on a sketch.
• Practise the two- and three-polaroid problems until the "halve first, then cos²θ" rule is automatic.
• Revise refraction and the refractive index in ray optics for NEET.

Next in the syllabus: dual nature, atoms and nuclei.