Question

A glass prism with a refracting angle of ${60}^{\circ }$ has a refractive index $1.52$ for red light. A parallel beam of white light is incident on one face of the prism at an angle of incidence, which gives minimum deviation for red light. Find the angle of minimum deviation for red light.
[Given: $\sin \left({50}^{\circ }\right)=0.760$]

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${50}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${40}^{\circ }$

Given that,
Angle of prism, $A={60}^{\circ }$
Refractive index for red light, ${\mu }_r=1.52$

We know,
$r_1+r_2=A$
$\Rightarrow r_1+r_2={60}^{\circ }$
Condition for minimum deviation is,
$r_1=r_2$
$\Rightarrow 2r_1={60}^{\circ }$
$\therefore r_1={30}^{\circ }$

Now applying Snell's law at first refracting surface,
${\mu }_{\text{air}}\sin i={\mu }_r\sin r_1$
$\Rightarrow \left(1\right)\left(\sin i\right)=\left(1.52\right)\sin {30}^{\circ }$
Or, $\sin i=0.76$
$\therefore i={\sin }^{-1}\left(0.76\right)$

Given, $\sin \left({50}^{\circ }\right)=0.76$
$\Rightarrow i={50}^{\circ }$

Again for red light,
${\delta }_{\text{min}}=(i+e)-A$
Or, ${\delta }_{\text{min}}=2i-A$
$\therefore {\delta }_{\text{min}}=(2\times {50}^{\circ })-{60}^{\circ }={40}^{\circ }$