A beam of light of wavelength `400 nm` is incident normally on a right angled prism as shown in Fig.
It is observed that light just grazes along the surface `AC` after falling on it. If refractive index `mu` of the material of prism varies with wavelength
`lamda` as `mu = 1.2 + (b)/(lamda^2)`
Calculate the value of `b and mu` of prism material for `lamda = 500 nm`. Given `theta = sin^-1 (0.625)`.
.

Explanation

As the ray goes grazingly along `AC, theta` must be the critical angle.
From `mu = (1)/(sin C) = (1)/(sin theta) = (1)/(0.625) = 1.6`
As `mu = 1.2 + (b)/(lamda^2)`
`1.6 = 1.2 + (b)/((400)^2`
`b = 0.4 xx (400)^2 = 64000 nm^2`
For `lamda = 500 nm`, the refractive index of the material of the prism is
`mu = 1.2 + (b)/(lamda^2) = 1.2 + (64000)/((500)^2) = 1.2 + 0.256`
`mu = 1.456`.