A light ray falls on a glass surface of refractive index $\sqrt{3}$, at an angle of $60^{\circ}$. The angle be

Question

A light ray falls on a glass surface of refractive index $\sqrt{3}$, at an angle of $60^{\circ}$. The angle between the refracted and reflected rays would be:

Accepted Answer

1. **Check for Brewster's Angle:** The relationship between the polarizing angle ($i_p$) and refractive index ($\mu$) is given by Brewster's Law: $	an i_p = \mu$.
2. **Verification:** Given $\mu = \sqrt{3}$ and angle of incidence $i = 60^{\circ}$.
   $$ 	an 60^{\circ} = \sqrt{3} $$
   Since the given angle of incidence satisfies the condition $	an i = \mu$, the incident angle is the Brewster's angle.
3. **Property:** At Brewster's angle, the reflected ray and the refracted ray are perpendicular to each other.
4. **Alternative Calculation:**
   - Angle of reflection $r' = i = 60^{\circ}$.
   - Angle of refraction $r$ from Snell's Law: $1 \cdot \sin 60^{\circ} = \sqrt{3} \sin r \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin r \Rightarrow \sin r = \frac{1}{2} \Rightarrow r = 30^{\circ}$.
   - Angle between rays $= 180^{\circ} - (r' + r) = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}$.

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