Question 27

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 O 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 O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $$\theta$$. The value of $$\sin \theta$$ is ________.

Correct Answer: 0.75

$$\tan \alpha = \sqrt{3} \implies \alpha = 60^\circ$$

$$\sqrt{3} \sin \beta = 1 \times \sin \alpha$$

$$\sqrt{3} \sin \beta = \sin 60^\circ = \frac{\sqrt{3}}{2} \implies \sin \beta = \frac{1}{2} \implies \beta = 30^\circ$$

Using the sine rule for the triangle formed inside the geometry:

$$\frac{R}{2 \sin 30^\circ} = \frac{x}{\sin 120^\circ}$$

$$\frac{R}{\sin 120^\circ} = \frac{R}{\frac{\sqrt{3}}{2} \times \sin \theta} \implies \sin \theta = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$$

$$\sin \theta = \frac{3}{4}$$

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