Question 30: Ray Optics

Question

A solid glass sphere of refractive index $n=\sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{R}{2}$, as shown in the figure. A very thin glass layer is present at the point 0 so that the air cavity (refractive index $n=1$ ) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point 0 and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $	heta$. The value of $\sin 	heta$ is ________.

Accepted Answer

The reflected light at O is "fully polarized," which means the incident angle at 0 is the Brewster angle. Brewster's angle occurs when the reflected and refracted rays are at $90^{\circ}$ to each other.
so Brewster angle is defined by
$$ 	an 	heta_B=\frac{n_2}{n_1} $$
So $	an \beta=\frac{\sqrt{3}}{1}$ (as at the point 0, light travels from air to glass so
$$ \begin{aligned} & 	ext { light, travels } \left.n_2=\sqrt{3}, n_1=1ight) \ & \Rightarrow \beta=60^{\circ} \end{aligned} $$
$$ 	ext { So } \angle D N O=60^{\circ}\left(\begin{array}{c} 	ext { As in } 	riangle D N O, DO = DN=\frac{R}{2} \ 	ext { so } \angle D O N=\angle D N O) \end{array}ight. $$
Now, using snell's law at point $N$,
$$ \begin{aligned} & n_1 \sin 	heta_1=n_2 \sin 	heta_2 \ \Rightarrow & \sqrt{3} \sin \alpha=1 	imes \sin 60^{\circ} \ \Rightarrow & \sqrt{3} \sin \alpha=\frac{\sqrt{3}}{2} \ \Rightarrow & \sin \alpha=\frac{1}{2} \Rightarrow \alpha=30^{\circ} \end{aligned} $$
In $	riangle D O N, \angle D\ O N+\angle O N D+\angle N D O=180^{\circ}$
$$ \begin{aligned} & 60^{\circ}+60^{\circ}+\angle N D O=180^{\circ} \ & \Rightarrow \angle N D O=60^{\circ} \end{aligned} $$
So $\angle C D N_{}=180^{\circ}-60^{\circ}=120^{\circ}$
$$ C N=2 C P $$
we can find $C P$ using $	riangle D P C$,
$$ \begin{aligned} \quad \cos 30^{\circ} & =\frac{C P}{C D} \ \Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{C P}{R / 2} \ \Rightarrow \quad C P & =\frac{\sqrt{3} R}{4} \ 	ext { So } \quad C N & =2\left(\frac{\sqrt{3} R}{4}ight)=\frac{\sqrt{3} R}{2} \end{aligned} $$
Now, using sine rule in $	riangle C N M$,
$$ \begin{aligned} & \frac{\sin 	heta}{C N}=\frac{\sin 120^{\circ}}{R} \ \Rightarrow \quad & \sin 	heta=\frac{\sqrt{3}}{2} B\left(\frac{\sqrt{3}, R}{2}ight) \ \Rightarrow \quad & \sin 	heta=\frac{3}{4}=0.75 \quad 	ext { Ans. } \end{aligned} $$

Ray Optics

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