Question

If refractive index of a glass prism is 1.5, then, find the angle of minimum deviation in relation with other terms of optics.

Answer 1

Dear student,
We know that
$$\begin{gathered} \mu =\frac{\sin \left({\displaystyle \frac{A+D}{2}}\right)}{\sin \left({\displaystyle \frac{A}{2}}\right)} \\ 1.5=\frac{\sin \left({\displaystyle \frac{60+D}{2}}\right)}{\sin \left({\displaystyle \frac{60}{2}}\right)} \\ 1.5\times .5=\sin \left(\frac{60+D}{2}\right) \\ \sin \left(\frac{60+D}{2}\right)=0.75 \\ \frac{60+D}{2}={\sin }^{-1}\left(0.75\right) \\ D={37}^0 \end{gathered}$$