Question

when refractive index is a linear function of x( like y=mx) how can we relate angle of incidence at any point to the value of slope?

Answer 1

Hmmm, not 100% sure what is being asked, but this may help...

Index of refraction is defined as
\[n = c/v\]
where c is the speed of light, and v is the speed in the given medium. For any transverse wave the speed is
\[v=\nu \lambda\]
where nu is the frequency and lambda is the wavelength. Now, recall that the energy of a wave depends on the frequency. By conservation of energy, the frequency must be in the same both inside and outside the medium. Therefore, since we know that the speed of light changes in a medium, this must mean that the wavelength must change. Throwing this into above yields
\[n = c/\nu_o \lambda\]
where nu_0 just indicates that it's the same on both sides...

Now look at Snell's Law
\[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\] ==>\[(c/\lambda_1) \sin(\theta_1) = (c/\lambda_2) \sin(\theta_2)\]
So I'm guessing that linear function of x means that the wavelength of light varies as one moves along the surface. The slope, is the "speed" at which it varies, which is constant.