Question

A prism of refractive index 1.53 is placed in water of refractive index 1.33. If the angle of the prism is 60${}^o$, calculate the angle of minimum deviation in water.

Answer 1

Minimum deviation $\delta =2i-A$

where A is prism angle $A={60}^o$

For minimum angle the angle of refraction should be half of prism angle.

So $r=A/2={30}^o$

now using the snell's law

$n_{water}sini=n_{glass}sinr$

$1.33sini=1.53sin30$

$i={35.11}^o$

Minimum deviation $\delta =2i-A$

$\delta =2\times 35.11-60={10.2}^o$