Question

A prism having refractive index $\mu =1.53$ is placed in water having refractive index $\mu =1.33$. If the angle of prism is ${60}^{\circ }$. Calculate the angle of minimum deviation in water.
$(\sin {35.1}^{\circ }=0.575)$

• A. ${30}^{\circ }$
• B. ${20.4}^{\circ }$
• C. ${10.2}^{\circ }$
• D. ${35.1}^{\circ }$

Answer 1

The correct option is C ${10.2}^{\circ }$

Given:
${\mu }_p=1.53;{\mu }_w=1.33;A={60}^{\circ }$

Using Snell's law,

${\mu }_w\sin i={\mu }_p\sin r$

$\Rightarrow \frac{\sin i}{\sin r}=\frac{{\mu }_p}{{\mu }_w}........\left(1\right)$

For minimum deviation, we know

$i=\left(\frac{{\delta }_{min}+A}{2}\right)\text{and}r=A/2$

Putting the value in eq. $\left(1\right)$, we get

$\frac{\sin (\frac{{\delta }_{min}+A}{2})}{\sin {30}^{\circ }}=\frac{1.53}{1.33}$

$\Rightarrow \sin \left(\frac{{\delta }_{min}+A}{2}\right)=1.15\times \frac{1}{2}=0.575$

$\Rightarrow \frac{{\delta }_{min}+A}{2}={\sin }^{-1}\left(0.575\right)={35.1}^{\circ }$

$\Rightarrow {\delta }_{min}=2\times 35.1-A={70.2}^{\circ }-{60}^{\circ }$

$\therefore {\delta }_{min}={10.2}^{\circ }$

Hence, option $\left(c\right)$ is correct.