Question

An equilateral prism has $\mu$=1.732. The angle of incidence for minimum deviation is

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. none of these

Answer 1

Answer: (A)

${60}^{\circ }$

For minimum deviation angle of incidence is equal to the angle of emergence
i.e, $r_1=r_2$
we know that $r_1+r_2=A$

so $r_1=r_2=\frac{A}{2}=30$

let $i$ be angle of incidence
then ,
${\mu }_{1}sin{\alpha }_1={\mu }_{2}sin{\alpha }_2$

$1\ast sini=1.732sin30$

$sini=\frac{1.732}{2}$

$i={60}^o$.