Question

A trihedral prism with refracting angle ${60}^{\circ }$ deviates a light ray by ${30}^{\circ }$. The refractive index of the material of prism

• A. may be equal to $\sqrt{2}$
• B. can not be greater than $\sqrt{2}$
• C. can not be less than $\sqrt{2}$
• D. None of these
  

Answer 1

Answer: (A), (B)

can not be greater than $\sqrt{2}$

Given :

$A={60}^{\circ }$ and $\delta ={30}^{\circ }$

We know that,
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(A/2\right)}$

$=\frac{\sin \left(\frac{{60}^{\circ }+{\delta }_m}{2}\right)}{\sin \left({60}^{\circ }/2\right)}$

$=\frac{\sin \left(\frac{{60}^{\circ }+{\delta }_m}{2}\right)}{\sin {30}^{\circ }}$

$=2\sin \left(\frac{{60}^{\circ }+{\delta }_m}{2}\right)$

Since one ray has been found out which has deviated by ${30}^{\circ }$, the angle of minimum deviation should be either equal or less than ${30}^{\circ }$, not more than ${30}^{\circ }$.

So, $\mu \le 2\sin \left(\frac{{60}^{\circ }+{30}^{\circ }}{2}\right)$

$\Rightarrow \mu \le \sqrt{2}$