Question

For glass prism $(\mu =\sqrt{3})$, the angle of minimum deviation is equal to the angle of prism. The angle of prism is:

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (C)

${60}^o$

$i_1+i_2=A+{\delta }_m,r_1+r_2=A$
Given, ${\delta }_m=A$
But in case of minimum deviation
$i_1=i_2=i,r_1=r_2=r\therefore r=A/2$
$\therefore \mu =\frac{sin\frac{a+{\delta }_m}{2}}{sinr}=\frac{sin\frac{A+A}{2}}{sin\frac{A}{2}}\mu =\frac{sinA}{sin\frac{A}{2}}$
$\mu =2cos\frac{A}{2},\sqrt{3}=2cos\frac{A}{2},\frac{\sqrt{3}}{2}=cos\frac{A}{2}$
$cos{30}^o=cos\frac{A}{2}\therefore A={60}^o$