Question

एक प्रिज्म का अपवर्तक कोण A = 60° तथा इसका अपवर्तनांक n = 3/2 है। न्यूनतम विचलन प्राप्त करने के
लिये आपतन कोण । कितना होगा ? न्यूनतम विचलन भी ज्ञात किजिये। [बाह्य माध्यम हवा (n = 1) लें]​

Answer 1

Answer:

please write in English I didn't understand your Question

Answer 2

The incidence angle is 48.59 degrees and angle of minimum deviation is 36.98 degrees.

Explanation:

Refracting angle o f a prism A=60∘ and its refractive index is n=3/2. What is angle of incidence i to get minimum deviation. Also, find the minimum deviation. Assume the surrounding medium to be air (n=1).

We have,

Angle of prism, A = 60

Refractive index, n = 3/2

For minimum deviation, the refractive index is given by :

$n=\dfrac{\sin (A+\delta _m)/2}{\sin A/2}$

$\delta_m$ is minimum deviation

$\dfrac{3}{2}=\dfrac{\sin (60+\delta _m)/2}{\sin 60/2}\\\\\dfrac{3}{2}=\dfrac{\sin (60+\delta _m)/2}{\sin (30)}\\\\\dfrac{3}{4}=\sin(\dfrac{60+\delta _m}{2})\\\\\sin^{-1}(\dfrac{3}{4})=\dfrac{60+\delta _m}{2}\ ........(1)\\\\48.49=\dfrac{60+\delta _m}{2}\\\\\delta_m=36.98^{\circ}$

We know that, angle of incidence is :

$i=\dfrac{A+\delta_m}{2}$

From equation (1) :

$i=\dfrac{A+\delta_m}{2}=\sin^{-1}(\dfrac{3}{4})\\\\i=48.59^{\circ}$

Hence, the incidence angle is 48.59 degrees and angle of minimum deviation is 36.98 degrees.

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Prism

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