Q 16. Find the equation of the curve that light traces as it passes from vacuum to a medium of variable refractive index as given in each sub-part.

The equation of the curve so formed can be found by solving the simple, first order separable linear ODE:

\[{dy \over dx} = \sqrt{ \left(\frac{\eta(y)}{\eta_{out} \sin i}\right) - 1}\]

Where, \(\eta(y)\) is the refractive index as a function of height \(y\), and \(i\) is the incident angle, a constant for a particular case. \(\eta_{out}\) is the (constant) refractive index of the medium from which light is incident onto the variable refractive index medium.

This can easily be derived by applying Snell’s law over an infinitesimal length.

(A): \[\eta=\eta_{_0} e^{\alpha y}\]

Here, \(\eta_{_0}\) and \(\alpha\) are constants that can be varied.

By integration, we arrive at:

\[x = \frac{1}{\alpha} \left( \tan^{-1} \left({\sqrt{\eta_{_0}^2 e^{2\alpha y} - \sin^2 i} \over \sin i } \right) - \tan^{-1} \left( {\sqrt{\eta_{_0}^2 - \sin^2 i} \over \sin i } \right) \right)\]

Graph(interactive):

(B): \[\eta = \eta_{_0} {1 \over y}\]

Here, \(\eta_{_0}\) is a constant.

We get,

\[(x \sin i - \eta_{_0})^2 + y^2 \sin^2 i=\eta_{_0}^2\]

Note that this is an equation of a circle. This indicates that the light ray undergoes TIR(Total Internal Reflection) at some point (the topmost point).

Graph(interactive):

(C): \[\eta = \eta_{_0} (y+\alpha)\]

Here, \(\eta_{_0}\) and \(\alpha\) are constants. We need \(\alpha\) here because we can’t have \(\eta\) less than 1(can you guess why?)

We get,

\[x = {\sin i \over \eta_{_0} } \ln\left| {\sqrt{\eta_{_0}^2 (y+\alpha)^2 - \sin^2 i} + \eta_{_0}(y+\alpha) \over \sqrt{\eta_{_0}^2 \alpha^2 - \sin^2 i} + \eta_{_0}\alpha} \right| \]

Graph(interactive):