A ray of light is incident on a glass slab at grazing incidence. The refractive index of the material of the slab is given by `mu=sqrt((1+y)` . If the thickness of the slab is `d=2m`, determing the equation of the trajectory of the ray inside the slab and the coordinates of the point where the ray exits from the slab. Take the origin to be at the point of entry of the ray.

Explanation

From the equation, we have
` (dy)/(dx) = cot ( theta_y) = sqrt(mu_(y)^(2) - sin^2 alpha)`
here ` mu= sqrt((1+y)) and alpha =90^@`
Therefore, `(dy)/(dx) = y^(1//2)`
Integrating with the boundary condition that y = 0 at x = 0, we get `y=(x^2)/( 4)` to be the equation of the path of the ray through the slab. The ray will obviously exit at the point `(2sqrt(2), d)`.