Question

A ray of light passing through a prism having refractive index `sqrt2` suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. What is the angle of prism?

Answer 1

`sqrt2=(sin i_1)/(sin(i_1//2))`
`=(2 sin(i_1//2)cos(i_1//2))/(sin i_1//2)`
Solving this, we get `i_1=90^@` and `r_1=i_1/2=45^@`
At minimum deviation,
`r_2=r_1=45^@`
`:. A=r_1+r_2=90^@`