Question

A ray of light travelling in air is incident at grazing
angle (Incident angle = 90${}^{\circ }$) on a long rectangular slab of
a transparent medium of thickness t = 1.0 m, The point of
incidence is the origin A(0,0). The medium has a variable
index of refraction n (y) given by
$n(y)=[k{y}^{3/2}+1{]}^{1/2}$ where k = 1.0$(m{)}^{-3/2}.$
The refractive index of air is 1.0.
(a) Obtain a relation between the slope of the trajectory of
the ray at a point B (x, y) in the medium and the incident
angle at that point.
(b) Obtain an equation for the trajectory y(x) of the ray in the medium.
(c) Determine the coordinates ($x_1,y_1$) of the point P, where
the ray intersects the upper surface of the slab-air
boundary.
(d) Indicate the path of the ray subsequently.

Answer 1

(a) Refractive index is a function of $y$. It varies along y-axis i.e. the boundary separating two media is parallel to $x$-axis or normal at any point will be parallel to $y$-axis.
Secondly, refractive index increases as $y$ is increased. Therefore, ray of light is travelling from rarer to denser medium i.e. it will bend towards the normal and shape of its trajectory will be as shown below.
Now, refer to figure (a)
Let $i$ be the angle of incidence at any point $B$ on its path
$\theta =90^{\circ }-i$ or $\tan \theta =\tan (90^{\circ }-i)=\cot i$
or slope $=\cot i$
(b) but tan $\theta =\frac{dy}{dx}$
$\Rightarrow \therefore \frac{dy}{dx}=\cot i$ ... (i)
Applying Snell's law at $A$ and $B$
$n_A\sin i_A=n_B\sin i_B\Rightarrow n_A=1$ because $y=0$
$\sin i_A=1$ because $i_A=90^{\circ }$ (Grazing incidence)
$n_B=\sqrt{k{y}^{3/2}+1}=\sqrt{y^{3/2}+1}$
because $k=1.0(m{)}^{-3/2}$
$\therefore (1)(1)=\sqrt{(y^{3/2}+1)}\sin i$
or $\sin i=\frac{1}{\sqrt{y^{3/2}+1}}$
$\therefore \cot i=\sqrt{y^{3/2}}$
or $y^{3/4}$ ... (ii)
Equating Eqs. (i) and (ii), we get
$\frac{dy}{dx}=y^{3/4}$ or $y^{-3/4}dy=dx$
or $\underset{0}{\overset{y}{\int }}{y}^{-3/4}dy=\underset{0}{\overset{x}{\int }}dx$
or $4y^{1/4}=x$... (iii)
The required equation of trajectory is $4y^{1/4}=x.$
(c) At point $P$, where the ray emerges from the slab
$y=1.0m$
$\therefore x=4.0m$ [From Eq. (iii)]
Therefore, coordinates of point P are
$P=(4.0m,1.0m)$
(d) As $n_A\sin i_A=n_P\sin i_P$ and as $n_A=n_P=1$
Therefore, $i_P=i_A=90^{\circ }$ i.e. the ray will emerge parallel to the boundary at $P$ i.e. at grazing emergence.