Question

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

• A. ${80}^0$
• B. ${70}^0$
• C. ${60}^0$
• D. ${50}^0$

Answer 1

Answer: (C)

${60}^0$

$\mu =1.732$
$D_m=A$
$\mu =\frac{sin\left(\frac{A+D_m}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$1.732=\frac{sin\left(\frac{2A}{2}\right)}{sin\left(\frac{A}{2}\right)}$
$1.732=\frac{2sin\frac{A}{2}\cdot cos\frac{A}{2}}{sin\left(\frac{A}{2}\right)}$
$cos\frac{A}{2}=\frac{1.732}{2}$
$\frac{A}{2}=co{s}^{-1}\left(0.866\right)$
$\frac{A}{2}={30}^0$ (or) $A={60}^0$