Question

What is the maximum value of angle of prism if it is equal to minimum deviation (M=3/2)

Answer 1

Answer on Question #52367-Physics-Optics

What is the maximum value of angle of prism if it is equal to minimum deviation $(\mu = \frac{3}{2})$

Solution

Angle of minimum deviation is $\delta_{m} = A$.

Angle of the prism is $A$.

Refractive index of prism is $\mu = \frac{3}{2}$.

The angle of deviation is related to refractive index as:

$$\mu = \frac {\sin \frac {(A + \delta_ {m})}{2}}{\sin \frac {A}{2}} = \frac {\sin \frac {(A + A)}{2}}{\sin \frac {A}{2}} = \frac {2 \sin \frac {A}{2} \cos \frac {A}{2}}{\sin \frac {A}{2}} = 2 \cos \frac {A}{2}.$$

Thus, the maximum value of angle of prism is

$$A = 2 \cos^ {- 1} \frac {\mu}{2} = 2 \cos^ {- 1} \frac {3}{4} = 82.82{}^{\circ}.$$

Answer: $82.82{}^{\circ}$.

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