Question

# The refracting angle of a prism is A and the refractive index is $$\displaystyle cot \left( \frac{A}{2} \right)$$. The angle of minimum deviation is

A
180oA
B
180o2A
C
180o3A
D
180o4A

Solution

## The correct option is B $$180^o-2A$$We know refractive index$$\displaystyle \mu= \frac{\displaystyle sin \left( \frac{A+\delta_m}{2} \right)}{\displaystyle sin \left( \frac{A}{2} \right)}$$Given, $$\displaystyle \mu = cot \left( \frac{A}{2} \right)$$Thus, $$\displaystyle cot \left( \frac{A}{2} \right) = \frac{\displaystyle sin \left( \frac{A+\delta_m}{2} \right)}{\displaystyle sin \left( \frac{A}{2} \right)}$$or $$\displaystyle cos \left( \frac{A}{2} \right) = sin \left( \frac{A+ \delta_m}{2} \right)$$$$\displaystyle sin \left( 90^o - \frac{A}{2} \right) = sin \left( \frac{A+ \delta_m}{2} \right)$$which gives $$\displaystyle \frac{A+ \delta_m}{2} =90^o - \frac{A}{2}$$or $$\displaystyle \delta_m = 180^o - 2A$$Physics

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