Q.
As shown in the figure, a light ray is incident on face $AB$ of prism $ABC$ . The second prism is then arranged in such a manner that emergent ray from prism $ABC$ is falling normally on face $A'B'$ of prism $A'B'C'$ . What is the net deviation produced by the optical system of the two prisms?
NTA AbhyasNTA Abhyas 2020
Solution:
From prism 1 :
$sini=\mu sinr\Rightarrow sin53^\circ =\frac{4}{3}\times sinr\Rightarrow sinr=\frac{3}{4}\times \frac{4}{5}=\frac{3}{5}\Rightarrow r=37^\circ Totaldeviation=16^\circ onbothends=32^\circ Inprism2:\frac{\delta_{1}}{A_{1}}=\frac{\delta_{2}}{A_{2}}\Rightarrow \delta_{2}=\frac{37}{74}\times 32^\circ =16^\circ Totaldeviation=\delta_{1}+\delta_{2}=32+16=48^\circ $
$sini=\mu sinr\Rightarrow sin53^\circ =\frac{4}{3}\times sinr\Rightarrow sinr=\frac{3}{4}\times \frac{4}{5}=\frac{3}{5}\Rightarrow r=37^\circ Totaldeviation=16^\circ onbothends=32^\circ Inprism2:\frac{\delta_{1}}{A_{1}}=\frac{\delta_{2}}{A_{2}}\Rightarrow \delta_{2}=\frac{37}{74}\times 32^\circ =16^\circ Totaldeviation=\delta_{1}+\delta_{2}=32+16=48^\circ $