1. Chapter at a Glance
Wave optics treats light as a wave. It explains effects that ray optics cannot: interference, diffraction and polarisation. This chapter covers Huygens' principle, coherent sources, Young's double-slit experiment (YDSE), single-slit diffraction and polarisation by polaroids and by reflection. NEET asks one or two questions here, most often on fringe width, intensity and Malus's law.
- Fringe width in YDSE and how it changes with λ, d, D and the medium.
- Conditions for bright and dark fringes; intensity at a point.
- Maximum and minimum intensities for unequal slits.
- Width of the central maximum in single-slit diffraction.
- Malus's law and Brewster's angle.
2. Huygens' Principle
- A wavefront is the locus of points vibrating in the same phase. Rays are perpendicular to wavefronts.
- A point source gives spherical wavefronts; a distant source gives plane wavefronts.
- Every point on a wavefront acts as a source of secondary wavelets. The new wavefront is the forward envelope of these wavelets.
- Huygens' construction gives both the laws of reflection and Snell's law, with n₁/n₂ = v₂/v₁.
- When light enters a denser medium, its speed and wavelength decrease (λ′ = λ/n) but its frequency stays the same.
3. Interference and Young's Double-Slit Experiment
Interference is the redistribution of light energy when two coherent waves overlap. Coherent sources have the same frequency and a constant phase difference. Two independent bulbs are never coherent, so YDSE uses two slits lit by the same source.
| Fringe | Path difference | Position on screen | NEET Note |
|---|---|---|---|
| Bright (n = 0, 1, 2…) | nλ | y = nλD/d | Central fringe is bright |
| Dark (n = 1, 2, 3…) | (2n − 1)λ/2 | y = (2n − 1)λD/2d | Halfway between bright fringes |
| Fringe width | — | β = λD/d | All fringes equally wide |
- Imax/Imin = (√I₁ + √I₂)²/(√I₁ − √I₂)². Intensity is proportional to slit width and to (amplitude)².
- In a medium of refractive index n, the fringe width becomes β/n.
- A thin sheet (thickness t, index n) over one slit shifts the whole pattern by (n − 1)tD/d towards that slit; the fringe width is unchanged.
- With white light, the central fringe is white and the others are coloured.
- Energy is conserved in interference; it is only redistributed.
4. Diffraction at a Single Slit
- Diffraction is the bending of light around obstacles or through apertures comparable in size to its wavelength.
- For a slit of width a, the first minima are at sin θ = ±λ/a, and the n-th minimum is at a sin θ = nλ.
- The central maximum has angular width 2λ/a and linear width 2λD/a. It is twice as wide as the secondary maxima, and much brighter.
- A narrower slit gives a wider central maximum. A longer wavelength (red) diffracts more than a shorter one (violet).
Interference comes from two separate coherent sources: its fringes are equally wide and equally bright. Diffraction comes from different parts of a single wavefront: the central fringe is widest and brightest, and the side fringes quickly fade.
5. Polarisation
- Only transverse waves can be polarised. Polarisation therefore proves that light is a transverse wave.
- Unpolarised light passing through a polaroid loses half its intensity: I = I₀/2, whatever the orientation.
- Two polaroids at θ: the second transmits I = (I₀/2)cos²θ. Crossed polaroids (θ = 90°) block all light.
- At Brewster's angle, the reflected light is completely polarised and the reflected and refracted rays are at 90° to each other (iB + r = 90°).
- Uses: polaroid sunglasses reduce glare, 3-D movie glasses, and photographic filters.
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Wavelength in a medium | λ′ = λ/n |
| Phase and path difference | φ = (2π/λ)Δx |
| Fringe width | β = λD/d |
| n-th bright fringe | y = nλD/d |
| Resultant intensity | I = I₁ + I₂ + 2√(I₁I₂)cos φ |
| Max / min intensity | (√I₁ ± √I₂)² |
| Shift due to a sheet | (n − 1)tD/d |
| Single-slit minima | a sin θ = nλ |
| Central maximum width | 2λD/a |
| Malus's law | I = I₀cos²θ |
| Brewster's law | tan iB = n |
7. Master Summary Table
| Phenomenon | Key idea | NEET Note |
|---|---|---|
| Huygens' principle | Secondary wavelets | Explains Snell's law |
| Interference | Coherent sources | Equal-width fringes |
| YDSE in a liquid | β → β/n | Fringes come closer |
| Thin sheet | Pattern shifts | β unchanged |
| Diffraction | Central max width 2λD/a | Narrow slit ⇒ wide pattern |
| Polaroid | I₀ → I₀/2 | Then cos²θ |
| Brewster angle | tan i = n | Reflected ⊥ refracted |
8. Mind Map
- Huygens
- Reflection, refraction
- Coherence
- YDSE, β = λD/d
- Intensity
- Single slit
- a sin θ = nλ
- Malus
- Brewster
- Transverse proof
9. Tips and Tricks
- Fringe width β ∝ λD/d: double D ⇒ double β; double d ⇒ half β.
- Slit width ratio = intensity ratio; amplitude ratio = √(intensity ratio).
- For coinciding bright fringes of λ₁ and λ₂: n₁λ₁ = n₂λ₂; take the smallest whole numbers.
- Three polaroids with the middle one at 45°: I = I₀/8 even when the outer two are crossed.
- Forgetting the first polaroid halves unpolarised light before applying cos²θ.
- Using 2λD/a (diffraction) for YDSE fringe width.
- Thinking a thin sheet changes the fringe width (it only shifts the pattern).
- Saying energy is destroyed at dark fringes.
10. Important One-liners
- Two independent sources of light cannot be coherent.
- Polarisation shows that light is a transverse wave; interference and diffraction show that it is a wave.
- The frequency of light does not change on refraction.
- In YDSE, the central fringe is bright when the slits are in phase.
- Sound waves cannot be polarised because they are longitudinal.
- At the polarising angle, the reflected and refracted rays are perpendicular.
- Diffraction is more noticeable for sound than for light because sound has a much longer wavelength.
11. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Click to reveal each answer and working.
Average level
In a YDSE, λ = 600 nm, the slit separation is 0.3 mm and the screen is 1.5 m away. Find the fringe width.
Show answer & solution
Concept: β = λD/d.
- β = 600 × 10⁻⁹ × 1.5/0.3 × 10⁻³.
- β = 9 × 10⁻⁷/3 × 10⁻⁴ = 3 × 10⁻³ m = 3 mm.
Answer: 3 mm
Find the Brewster angle for a glass surface of refractive index 1.6, and the angle of refraction at that incidence.
Show answer & solution
Concept: tan iB = n; iB + r = 90°.
- iB = tan⁻¹(1.6) ≈ 58°.
- r = 90° − 58° = 32°.
Answer: about 58°; about 32°
Unpolarised light of intensity I₀ passes through a polariser and then an analyser whose axis is at 30° to the polariser. Find the final intensity.
Show answer & solution
Concept: first polaroid gives I₀/2; then Malus's law.
- After the polariser: I₀/2.
- After the analyser: (I₀/2)cos²30° = (I₀/2)(3/4).
- I = 3I₀/8.
Answer: 3I₀/8
Light of wavelength 600 nm in air enters glass of refractive index 1.5. Find its wavelength and frequency in glass.
Show answer & solution
Concept: frequency is unchanged; λ′ = λ/n.
- λ′ = 600/1.5 = 400 nm.
- f = c/λ = 3 × 10⁸/600 × 10⁻⁹ = 5 × 10¹⁴ Hz, the same in glass.
Answer: 400 nm; 5 × 10¹⁴ Hz
Light of wavelength 500 nm falls on a slit 0.2 mm wide. Find the width of the central maximum on a screen 2 m away.
Show answer & solution
Concept: width = 2λD/a.
- Width = 2 × 500 × 10⁻⁹ × 2/0.2 × 10⁻³.
- = 2 × 10⁻⁶/2 × 10⁻⁴ = 10⁻² m = 1 cm.
Answer: 1 cm
Advanced level
A YDSE gives a fringe width of 3 mm in air. The whole apparatus is immersed in water (n = 4/3). Find the new fringe width and explain the change.
Show answer & solution
Concept: λ in water = λ/n, so β also falls by the factor n.
- β′ = β/n = 3/(4/3).
- β′ = 2.25 mm; the fringes come closer because the wavelength is shorter in water.
Answer: 2.25 mm
The widths of the two slits in a YDSE are in the ratio 4 : 1. Find the ratio of maximum to minimum intensity in the pattern.
Show answer & solution
Concept: I ∝ width; amplitude ∝ √I.
- I₁ : I₂ = 4 : 1 ⇒ a₁ : a₂ = 2 : 1.
- Imax ∝ (2 + 1)² = 9; Imin ∝ (2 − 1)² = 1.
- Imax : Imin = 9 : 1.
Answer: 9 : 1
Two coherent waves, each of intensity I₀, meet with a phase difference of π/3. Find the resultant intensity. For λ = 600 nm, what path difference produces this phase difference?
Show answer & solution
Concept: I = 4I₀cos²(φ/2); Δx = (λ/2π)φ.
- I = 4I₀cos²(π/6) = 4I₀ × 3/4 = 3I₀.
- Δx = (600/2π) × (π/3) = 100 nm.
Answer: 3I₀; 100 nm
A YDSE (d = 2 mm, D = 1.2 m) is lit by 650 nm and 520 nm light together. Find the least distance from the central maximum at which bright fringes of the two wavelengths coincide.
Show answer & solution
Concept: coincidence when n₁λ₁ = n₂λ₂.
- n₁ × 650 = n₂ × 520 ⇒ n₁/n₂ = 520/650 = 4/5.
- Smallest: n₁ = 4 (for 650 nm), n₂ = 5 (for 520 nm); common path difference 2600 nm.
- y = 4 × 650 × 10⁻⁹ × 1.2/2 × 10⁻³ = 1.56 × 10⁻³ m.
Answer: 1.56 mm
In a YDSE with light of wavelength 540 nm, a thin transparent sheet of refractive index 1.6 placed over one slit shifts the central bright fringe by 4 fringe widths. Find the thickness of the sheet.
Show answer & solution
Concept: extra optical path (n − 1)t equals the number of fringes shifted × λ.
- (n − 1)t = 4λ.
- t = 4 × 540 × 10⁻⁹/0.6.
- t = 3.6 × 10⁻⁶ m = 3.6 μm. The pattern shifts towards the covered slit.
Answer: 3.6 μm
12. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Which phenomenon proves that light waves are transverse?
- Interference
- Diffraction
- Polarisation
- Reflection
Show solution
Only transverse waves can be polarised. Interference and diffraction happen for all waves, including longitudinal sound waves, so they show only that light is a wave.
Answer: (C)
In a YDSE, if the distance between the slits is doubled, the fringe width:
- Doubles
- Halves
- Is unchanged
- Becomes four times
Show solution
β = λD/d ∝ 1/d, so doubling d halves β. Doubling D would double β instead.
Answer: (B)
Unpolarised light of intensity 32 W m⁻² passes through two polaroids whose axes are at 60°. The emerging intensity is:
- 16 W m⁻²
- 8 W m⁻²
- 4 W m⁻²
- 2 W m⁻²
Show solution
After the first polaroid: 16 W m⁻². After the second: 16 × cos²60° = 16 × 1/4 = 4 W m⁻². (B) applies cos 60° instead of cos², and (A) stops after the first polaroid.
Answer: (C)
In single-slit diffraction, if the slit is made narrower, the central maximum:
- Becomes narrower
- Becomes wider
- Is unchanged
- Disappears
Show solution
The width is 2λD/a ∝ 1/a, so a narrower slit spreads the light more. The central maximum also becomes dimmer, because less light gets through.
Answer: (B)
Two sources are coherent if they have:
- The same amplitude
- The same frequency and a constant phase difference
- The same intensity
- Different frequencies
Show solution
Coherence means a constant phase relationship, which needs the same frequency. Equal amplitudes improve the contrast of the fringes, but they are not required for coherence.
Answer: (B)
At Brewster's angle of incidence, the angle between the reflected and refracted rays is:
- 0°
- 45°
- 90°
- 180°
Show solution
At iB, tan iB = n gives iB + r = 90°. This makes the reflected and refracted rays perpendicular to each other.
Answer: (C)
In a YDSE, the intensity at the central maximum is I₀. The intensity at a point where the path difference is λ/4 is:
- I₀
- I₀/2
- I₀/4
- Zero
Show solution
φ = (2π/λ)(λ/4) = π/2. I = I₀cos²(φ/2) = I₀cos²(π/4) = I₀/2. Here I₀ = 4 × (the intensity of one slit). The intensity is zero only for a path difference of λ/2.
Answer: (B)
A thin glass sheet is placed in front of one slit in a YDSE. Which statement is correct?
- The fringe width increases
- The fringe pattern shifts towards the covered slit, and the fringe width is unchanged
- The fringes disappear
- The pattern shifts away from the covered slit
Show solution
The sheet adds an optical path (n − 1)t to that slit's light. The point of zero path difference therefore moves towards the covered slit. Since β = λD/d does not involve t, the fringe width does not change.
Answer: (B)
13. Quick Revision Summary
- Huygens: secondary wavelets; λ′ = λ/n, f unchanged.
- YDSE: β = λD/d; bright nλ, dark (2n − 1)λ/2; in liquid β/n.
- I = 4I₀cos²(φ/2); Imax/Imin = (a₁ + a₂)²/(a₁ − a₂)².
- Single slit: a sin θ = nλ (minima); central max 2λD/a.
- Polaroid I₀/2 then cos²θ; Brewster tan i = n.