Question

A light ray is incident on a glass sphere at an angle of incidence ${60}^{\circ }$ as shown in the figure. The amount of total deviation suffered by the light ray when it comes out of the glass sphere will be

• A. ${45}^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Applying Snell's law at the interface where first refraction occurs.
${\mu }_1\sin i={\mu }_2\sin r$
$\Rightarrow {\mu }_a\sin {60}^{\circ }={\mu }_g\sin r$
$\Rightarrow 1\times \frac{\sqrt{3}}{2}=\sqrt{3}\times \sin r$
$\Rightarrow \sin r=\frac{1}{2}$
$\Rightarrow r={30}^{\circ }$
Here, the line joining point of incidence and centre of sphere is normal $\because CM=CN=$ radius.


From symmetry, $r=r_1={30}^{\circ }$
Applying Snell's law at interface during second refraction,
${\mu }_g\sin r_1={\mu }_a\sin \theta$
$\Rightarrow \sqrt{3}\sin {30}^{\circ }=1\times \sin \theta$
$\Rightarrow \theta ={60}^{\circ }$
Deviation during first refraction, $\delta {\theta }_1={60}^{\circ }-{30}^{\circ }={30}^{\circ }$
Deviation during second refraction,
$\delta {\theta }_2={60}^{\circ }-{30}^{\circ }={30}^{\circ }$
$\therefore {\delta }_{total}={30}^{\circ }+{30}^{\circ }={60}^{\circ }$
$[$clockwise$]$

Why this question? It challenges you for application of Snell's law at the spherical interface. Tips: The normal direction for a spherical surface passes through its centre.