Question

a beam of monochromatic light is incident at angle of incidence 50 degree on one face of an equilateral prism the angle of emergence is 40 Degrees then the angle of minimum deviation is

Answer 1

Answer: The angle of minimum deviation is $40\degree$.

Explanation:

The expression for the angle of deviation is as follows,

$i+e=A-\delta$                                                              ...... (1)

Here, i is the angle of incidence, e is the angle of emergence, A is the angle of prism and $\delta$ is the angle of deviation.

For minimum angle of deviation $\delta _{m}$,

i=e

Put e=i in the equation (1).

$2i=A-\delta _{m}$  

Put $i=50\degree$ and $e=40\degree$.

$2(50\degree)=60\degree-\delta _{m}$

$2(50\degree)-60\degree=\delta _{m} \\\delta _{m}=40\degree$

Therefore, the value of the angle of minimum deviation is $40\degree$.