Question

the angle of the prism is 60 degrees and its refractive index is root 2 . at minimum deviation conditions what is the possible value of the angle of incidence?
can anyone solve this with steps?!

Answer 1

Answer:

The required angle of incidence is 45°

Step-by-step explanation:

Given :

Angle of the prism, A = 60°

Refractive index, μ = √2

To find :

the angle of incidence at minimum deviation

Solution :

The refractive index of the prism is given by,

  $\longmapsto \sf \mu=\dfrac{\sin\bigg(\dfrac{A+D}{2}\bigg)}{\sin \dfrac{A}{2}}$

where D is the minimum deviation angle.

At minimum deviation, the angle of incidence (i) is equal to the angle of emergence (e).

i = e

=> The angle of deviation is given by,

  D = i + e - A

    i + e = A + D

    i + i = A + D

   2i = A + D

    i = (A + D)/2

Substitute the given values,

$\rm \sqrt{2}=\dfrac{\sin i}{\sin \dfrac{60^{\circ}}{2}} \\\\ \rm \sqrt{2}=\dfrac{\sin i}{\sin 30^{\circ}} \\\\ \rm \sin i=\sqrt{2} \times \dfrac{1}{2} \\\\ \rm \sin i=\sqrt{2} \times \dfrac{1}{\sqrt{2}^2} \\\\ \rm \sin i=\dfrac{1}{\sqrt{2}} = \sin 45^{\circ} \\\\ \longmapsto \boxed{\rm i=45^{\circ}}$

Therefore, the angle of incidence at minimum deviation is 45°