Question

For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of prism. Then the angle of the prism is :

• A. $50$
• B. ${60}^o$
• C. ${70}^o$
• D. None of these

Answer 1

The correct option is B ${60}^o$

Given, refractive index of prism $=1.732$ Let the angle of prism is A.

The angle of minimum deviation

= Total angle of prism

${\delta }_m=A$

The refractive index of prism

$mu=\frac{sin\left[\frac{A+{\delta }_m}{2}\right]}{sin\frac{A}{2}}$

or $mu=\frac{sin\left[\frac{A+A}{2}\right]}{sin\frac{A}{2}}$

or $1.732=\frac{sinA}{sin\frac{A}{2}}$

or $\sqrt{3}=\frac{2sin\frac{A}{2}\cdot cos\frac{A}{2}}{sin\frac{A}{2}}$

or $A={60}^o$