We have
`(dy)/(dx)=cot(theta_(y))=sqrt(mu_(y)^(2)-sin^(2)theta)/(sintheta)`
Here, `mu=sqrt((1+y))` and `theta=90^(@)` Therefore, `(dy)/(dt)=y^(1//2)`
Integratin with the boundary condition that `y=0` at `zx=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 `(2sqrt2m,2m)`.