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  • For a glass prism mu=sqrt{3} the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

For a glass prism \mu=\sqrt{3} the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Answers (1)

The refractive index of prism angle A and angle of minimum deviation \delta _{m} is given by

\mu=\frac{\sin\left [ \frac{\left ( A+\delta _{m} \right )}{2} \right ]}{\sin\left ( \frac{A}{2} \right )}

Here we are given, \delta _{m}=A

Substituting the value, we have

\mu=\frac{\sin A}{\sin \frac{A}{2}}

\Rightarrow \mu=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}}=2 \cos\frac{A}{2}

\Rightarrow \mu=\frac{\sin A}{\sin \frac{A}{2}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}}=2 \cos\frac{A}{2}

For the given value of refractive index,

We have, \cos\frac{A}{2}=\frac{\sqrt{3}}{2}\Rightarrow \frac{A}{2}=30^{o}

Or A=60^{o}

This is the required value of prism angle.

Posted by

infoexpert23

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