Question

If the refractive index of the material of a prism is $\sqrt{3}$, then, the angle of minimum deviation of the prism is:

(Take angle of prism to be ${60}^o$)

• A. Equal to ${45}^o$
• B. Less than ${45}^o$
• C. Greater than ${45}^o$
• D. Data insufficient

Answer 1

The correct option is C Greater than ${45}^o$

Refractive index of the material of a prism
$\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
Where, $\mu$ s refractive index, A is angle of prism, ${\delta }_m$ is angle of minimum deviation.
Given, $\mu =\sqrt{3},A={60}^o$
$\sqrt{3}=\frac{sin\left(\frac{{60}^o+{\delta }_m}{2}\right)}{sin\left(\frac{{60}^o}{2}\right)}$
$\frac{\sqrt{3}}{2}=sin({30}^o+\frac{{\delta }_m}{2})$
$\Rightarrow {\delta }_m={60}^o$
We can see that the angle of minimum deviation of the prism is greater than ${45}^o$.