Question

The angle of minimum deviation produced by a 60${}^o$ prism is 40${}^o$. Calculate the refractive index of the material of the prism.

Answer 1

Minimum deviation $\delta =2i-A$

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

So $\delta ={40}^o$

we know

$\delta =2i-A$

$40=2i-60$

$i={50}^o$

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

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

$sin50=n_{glass}sin30$

$n_{glass}=1.53$ ,