Question

Find the angle of deviation suffered by the light ray shown in figure. The refractive index $\mu =1.5$ for the prism material.

Answer 1

Given $\mu =1.5$

And angle of prism $=4^{\circ }$

$\therefore \mu =\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin A/2}=\frac{(A+{\delta }_m)/2}{\left(A/2\right)}$ (for small angle $\sin \theta =\theta$)

$\Rightarrow \mu =\frac{A+{\delta }_m}{2}\Rightarrow 1.5=\frac{4^{\circ }+{\delta }_m}{4^{\circ }}\Rightarrow {\delta }_m=4^{\circ }\times \left(1.5\right)-4^{\circ }=2^{\circ }$