Question

A prism of refractive index $\sqrt{2}$ has refracting angle ${60}^0$. Angle of maximum deviation is:

• A. ${45}^0$
• B. $si{n}^{-1}\left(\sqrt{2}sin{15}^0\right)$
• C. ${30}^0+si{n}^{-1}\left(\sqrt{2}sin{15}^0\right)$
• D. none

Answer 1

Answer: (C)

${30}^0+si{n}^{-1}\left(\sqrt{2}sin{15}^0\right)$

Maximum deviation takes place.

when the grazing incidence emergence happens.

Deviation will be same in incidence and grazing emergence. and Deviation will be maximum

${\mu }_1\sin i={\mu }_2\sin \gamma atAB$

$\left(1\right)\sin \left(70\right)=\sqrt{2}\sin C$

$C={45}^{\circ }$

so ${\gamma }_2=60-45={15}^{\circ }$

$\begin{array}{cc}{\mu }_1\sin i& ={\mu }_2\sin \gamma \text{at}BC\\ \sqrt{2}\sin 15& =\left(1\right)\text{sine.}e={\sin }^{-1}\sqrt{2}\sin 15\\ \text{Deviation}=& +(180-(90+c\left)\right)+{\sin }^{-1}\left(\sqrt{2}\sin 15\right)-15\\ & =90-45-15+{\sin }^{-1}\left(\sqrt{2}\sin 15\right)\\ & =30+{\sin }^{-1}\left(\sqrt{2}\sin 15\right)\end{array}$

So maximum deviation $=30+{\sin }^{-1}\left(\sqrt{2}\sin 15\right)$

Hence option $c$ is correct