Question

The angle of incidence for a ray of light at a refracting surface of a prism is 45o. The angle of prism is 60o. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are :​

Answer 1

GIVEN :

• Angle of Incidence (i) = 45°
• Angle of Prism (A) = 60°
• Minimum deviation takes place.

TO FIND :

• Angle of minimum deviation
• Refractive index of the material

ANSWER :

We know, for minimum deviation

Angle of incidence (i) = Angle of emergence (e)

$\sf \therefore$ i = e = 45°

We have, for Angle of Deviation

$\boxed { \red{ \sf \: \delta = (i \: + \: e) - \:A }}$

$\implies \: \sf \: \delta = ( {45}^{o} \: + \: {45}^{o} ) - \: {60}^{o}$

$\implies \: \sf \: \delta = {90}^{o} \: - \: {60}^{o}$

$\implies \: \sf \: \delta _{m} = {30}^{o}$

Therefore, Angle of minimum deviation is 30°

Now, lets find out the refractive index of the prism using the formula,

$\boxed {\blue{ \sf \mu \: = \frac{sin ( \frac{ A + \: \delta _{m}}{2}) }{sin \: \frac{ A }{2}}}}$

$\implies \: \sf \: \mu \: = \frac{sin ( \frac{ {60}^{ \: o} + \: {30}^{o} }{2}) }{sin \: \frac{ {60}^{ \: o} }{2}}$

$\implies \: \sf \: \mu \: = \frac{sin ( \frac{ 90}{2}) }{sin \: 30}$

$\implies \: \sf \: \mu \: = \frac{sin \ \: 45 }{sin \: 30}$

$\implies \: \sf \: \mu \: = \frac{ \frac{1}{ \sqrt{2} } }{ \frac{1}{2}}$

$\boxed{\sf \: \therefore \: \mu \: = \sqrt{2} }$

So, the refractive index of the medium of the prism is √2.

Answer 2

Answer:

Angle of minimum deviation: $\sf \delta _m = 30^\circ$

Refractive index of the material of the prism: $\sf \mu = \sqrt{2}$

Explanation:

Angle of prism: A = 60°

Angle of incidence of a ray of light at a refracting surface of prism: i = 45°

In case of minimum deviation:

$\boxed{\bold{\rm \delta _m = 2i - A}}$

$\therefore$

$\implies \sf \delta _m = 2 \times 45^\circ - 60^\circ \\ \\ \implies \sf \delta _m = 90^\circ - 60^\circ \\ \\ \implies \sf \delta _m = 30^\circ$

So, Angle of minimum deviation: $\sf \bold{\delta _m = 30^\circ}$

Formula of Refractive Index of Material of Prism for Minimum Deviation (Prism Formula):

$\boxed{\bold{\rm \mu = \dfrac{sin \bigg(\dfrac{A + \delta _m}{2} \bigg)}{sin \dfrac{A}{2}} }}$

By substituting value in the formula we get:

$\implies \sf \mu = \dfrac{sin \bigg( \dfrac{60^\circ +30^\circ}{2} \bigg)}{sin \dfrac{60^\circ}{2}} \\ \\ \implies \sf \mu = \dfrac{sin \bigg( \dfrac{90^\circ}{2} \bigg)}{sin 30^\circ} \\ \\ \implies \sf \mu = \dfrac{sin 45^\circ}{sin30^\circ} \\ \\ \implies \sf \mu = \dfrac{ \dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \\ \\ \implies \sf \mu = \dfrac{2}{\sqrt{2}} \\ \\ \implies \sf \mu = \sqrt{2}$

So, Refractive index of the material of the prism: $\sf \bold{\mu = \sqrt{2}}$