Question

The refractive index for the material of a ${60}^{\circ }$ prism is 1.50. Then the angle of incidence for minimum deviation is nearly.$\left(\sin {42}^{\circ }\approx \frac{2}{3}\text{and}\sin {49}^{\circ }\approx \frac{3}{4}\right)$

• A. ${30}^o$'
• B. ${49}^o$'
• C. ${38}^o$'
• D. ${28}^o$'

Answer 1

The correct option is A ${30}^o$'

Given that,

Redfractive index, $n=1.5$

Using the relation,

$\mu =\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$1.5=\frac{\sin \left(\frac{60+D_m}{2}\right)}{\sin 30}$

$\sin \left(\frac{60+D_m}{2}\right)=\frac{1}{\sqrt{2}}=\sin 45$

$\left(\frac{60+D_m}{2}\right)=45$

$D_m={30}^0$

the angle of incidence for minimum deviation is 30⁰