Question

Question #ea4f0

Answer 1

We know the following relation of refractive index (`mu`) of the prism with the angle of prism (A) and minimum deviation (`del_m`)

`mu=sin((del_m+A)/2)/sin(A/2)`

As per given condition we have `del_m=A`

So we have

`mu=sin((A+A)/2)/sin(A/2)`

`=>mu=sinA/sin(A/2)`

`=>mu=(2sin(A/2)cos(A/2))/sin(A/2)`

`=>cos(A/2)=mu/2`

`=>A=2cos^-1(mu/2)`

If we take for glass `mu=1.5`

Then `A=2cos^-1(1.5/2)=82.8^@`

If we think otherwise

This relation reveals that the value of `A` increases with decrease in value of `mu` and A will be maximum when `mu` is minimum.The medium of the material of the prism is optically denser than the surrounding medium. So the minimum value of the refractive (`mu`) of the medium with respect to the surrounding one may be taken as `~~1`.

Imposing this condition we can have the maximum value of `A`.

Hence `" "A_"max"=2cos^-1(1/2)=2xx60^@=120^@`