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Unpolarised light is incident on a medium with refractive index $\sqrt{3}$. The angle of incidence for which reflected light is completely plane polarised is:
Options
1
$30°$
2
$45°$
3
$60°$
4
$75°$
Correct Answer
$60°$
Solution
1

Brewsters law: $\tan\theta_B = n$

$$\tan\theta_B = \sqrt{3} \implies \theta_B = \tan^{-1}(\sqrt{3}) = 60°$$
2

At $\theta_B = 60°$: reflected ray is completely plane polarised

Reflected and refracted rays are perpendicular: $\theta_r = 90° - 60° = 30°$

Brewster: $\tan\theta_B = n = \sqrt{3}$ $\Rightarrow$ $\theta_B = 60°$
Reflected light is completely polarised at this angle
Theory: Optics
1. Polarisation of Light

Light is a transverse wave — electric field oscillates perpendicular to direction of propagation. Unpolarised light: electric field oscillates in all directions. Plane polarised: oscillates in one plane only. Methods of polarisation: (1) Selective absorption (Polaroids/dichroic crystals). (2) Reflection at Brewster angle. (3) Scattering (blue sky). (4) Double refraction (birefringent crystals like calcite). Malus law: $I = I_0\cos^2\theta$ (intensity after analyser at angle $\theta$ to polariser).

2. Brewsters Law

When light reflects off a dielectric surface at Brewster angle $\theta_B$: reflected ray is completely polarised with E-field perpendicular to plane of incidence. $\tan\theta_B = n_2/n_1$ (from medium 1 to medium 2). For air-glass ($n=1.5$): $\theta_B = \tan^{-1}(1.5) = 56.3°$. At Brewster angle: reflected + refracted rays are perpendicular ($\theta_B + \theta_r = 90°$). p-polarisation (parallel) has zero reflection coefficient. Only s-polarisation (perpendicular) reflected.

3. Malus Law

$I = I_0\cos^2\theta$ where $\theta$ = angle between plane of polarisation and transmission axis of analyser. $\theta=0°$: $I = I_0$ (maximum). $\theta=90°$: $I=0$ (complete extinction). For unpolarised light through polariser: $I = I_0/2$ (average over all $\theta$). Applications: LCD screens (two polaroids with liquid crystal between them). 3D cinema: different polarizations for left and right eyes. Photography: polaroid filter reduces glare (reflected polarised light blocked).

4. Double Refraction (Birefringence)

Certain crystals (calcite, quartz) have two different refractive indices. Ordinary ray (o-ray): obeys Snell law, polarised perpendicular to optic axis. Extraordinary ray (e-ray): does not obey Snell law, polarised parallel to optic axis. Two rays travel at different speeds → phase difference → can produce circularly/elliptically polarised light. Quarter-wave plate: path difference $= \lambda/4$ → circular polarisation (from linear). Half-wave plate: path difference $= \lambda/2$ → rotates polarisation by $2\theta$ (where $\theta$ = angle of incidence to optic axis).

5. Scattering and Polarisation

Rayleigh scattering: $I \propto 1/\lambda^4$. Blue light scattered more than red → blue sky, red sunset. Scattered light from sky is partially polarised. Maximum polarisation at 90° to Sun direction. Bees and insects use sky polarisation for navigation. Light scattered perpendicular to incident direction is completely polarised (Tyndall effect). Optical activity: some molecules (glucose, fructose) rotate plane of polarised light. Measurement of optical rotation: polarimetry (used to determine sugar concentration).

6. Total Internal Reflection

When light goes from denser to rarer medium at angle $> $ critical angle $\theta_c$: all light reflected. $\sin\theta_c = n_{rarer}/n_{denser} = 1/n$ (air-glass). For glass ($n=1.5$): $\theta_c = 41.8°$. Applications: optical fibres (light trapped by TIR, bandwidth $>$ 1 THz), diamonds (cut to maximise TIR → brilliance, critical angle $24.4°$), mirage (hot air = rarer medium), prism binoculars (two TIR prisms fold path). Frustrated TIR: if two glass surfaces brought close together, light tunnels through the gap (evanescent wave).

7. Interference of Light

Youngs double slit: fringe width $\beta = \lambda D/d$ where $D$ = screen distance, $d$ = slit separation. Bright fringes at path difference $= m\lambda$. Dark at $(2m+1)\lambda/2$. Coherent sources required (stable phase relationship). Thin film interference: $2nt\cos r = m\lambda$ (for dark fringes, with phase reversal). Anti-reflection coating: $t = \lambda/(4n)$. Newton rings: circular interference fringes. Radius of $m$th ring: $r_m = \sqrt{m\lambda R}$ where $R$ = radius of curvature of lens.

8. Diffraction

Single slit: first minimum at $\sin\theta = \lambda/a$ (where $a$ = slit width). Central maximum width $= 2\lambda D/a$. Diffraction grating: $d\sin\theta = m\lambda$. Resolving power $= mN$ ($N$ = number of slits). Rayleigh criterion: just resolved when central max of one falls on first min of other. $\theta_{min} = 1.22\lambda/D$ for circular aperture. Telescope resolving power: $D/1.22\lambda$ (larger mirror/lens = better resolution). Bragg diffraction: $2d\sin\theta = n\lambda$ (X-rays by crystals). Used to determine crystal structures.

Frequently Asked Questions
1. Why does n = sqrt(3) give theta_B = 60°? ⌄
$\tan\theta_B = n = \sqrt{3}$. We know $\tan(60°) = \sqrt{3}$. Therefore $\theta_B = 60°$. Important $\tan$ values: $\tan(30°) = 1/\sqrt{3} \approx 0.577$, $\tan(45°) = 1$, $\tan(60°) = \sqrt{3} \approx 1.732$. For glass with $n=1.5$: $\theta_B = \tan^{-1}(1.5) = 56.3°$. For water ($n=1.33$): $\theta_B = 53.1°$. For diamond ($n=2.42$): $\theta_B = 67.5°$.
2. At Brewster angle, what is the angle of refraction? ⌄
At Brewster angle: $\theta_B + \theta_r = 90°$ always (this is actually what defines Brewster angle from Fresnel equations). Here $\theta_r = 90° - 60° = 30°$. Verification: Snell law: $n_1\sin\theta_B = n_2\sin\theta_r$. $1\times\sin(60°) = \sqrt{3}\times\sin(30°)$. LHS: $\sqrt{3}/2$. RHS: $\sqrt{3}\times 1/2 = \sqrt{3}/2$. ✓ Consistent.
3. Why is only the perpendicular component reflected at Brewster angle? ⌄
From Fresnel equations: reflection coefficient for p-polarization (parallel to plane of incidence): $r_p = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}$. At Brewster angle: $\theta_i + \theta_t = 90°$, so $\theta_t = 90° - \theta_i$. Then $\cos\theta_t = \cos(90°-\theta_i) = \sin\theta_i$ and $\cos\theta_i = \sin\theta_t$. Substituting: $r_p = \frac{n_2\sin\theta_t - n_1\sin\theta_i}{n_2\sin\theta_t + n_1\sin\theta_i} = 0$ (using Snell: $n_1\sin\theta_i = n_2\sin\theta_t$). Hence $r_p = 0$ → no reflected p-polarization.
4. How is polarisation by reflection used practically? ⌄
Photography: polarising filter on camera lens reduces reflections from water, glass, roads (reflected light is polarised). Rotate filter to block polarised reflected light → reduce glare, increase colour saturation. Sunglasses: Polaroid lenses with vertical transmission axis block horizontally polarised glare from flat surfaces (water, roads). Anti-reflection in optics: laser windows at Brewster angle (Brewster window) transmit p-polarised light with zero reflection. Used in laser cavities to output polarised beam.
5. What is the difference between polariser and analyser? ⌄
Both are polaroid sheets with a specific transmission axis. Polariser: first sheet that converts unpolarised light to plane-polarised light. Transmits $I_0/2$ (half the original intensity). Analyser: second sheet that analyses (tests) the polarisation. If at angle $\theta$ to polariser: transmits $I_0/2 \times \cos^2\theta$ (Malus law). At $\theta=0°$: maximum transmission. At $\theta=90°$ (crossed polaroids): zero transmission. There is no physical difference between polariser and analyser — both are identical polaroid sheets. The naming depends on which is first in the light path.
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