Question

A beam of monochromatic light is incident at $i={50}^o$ on one face of an equilateral prism, the angle of emergence is ${40}^o$, then the angle of minimum deviation is:

• A. ${30}^o$
• B. $<{30}^o$
• C. $\le {30}^o$
• D. $\ge {30}^o$

Answer 1

The correct option is D $\ge {30}^o$

We know that the angle of deviation is given by

$\Rightarrow i+e=A-\delta$

where, $i\to angleofincidence$

$e\to angleofemergence$

$A\to angleofprism$

$\delta \to angleofdeviation$

For angle of minimum deviation ${\delta }_m,i=e$

Thus
$2i=A-{\delta }_m$ $\left(1\right)$

$\Rightarrow A={60}^0$ (since its a equilateral prism)

$\Rightarrow i={50}^0$ (given)

Substituting these values in $\left(1\right)$ we get,

$\Rightarrow 2\left({50}^0\right)={60}^0-{\delta }_m$

$\Rightarrow {100}^0={60}^0-{\delta }_m$

On solving we get ${\delta }_m={40}^0$

Thus minimum angle of deviation ${\delta }_m={40}^0$