Question

A ray of light travelling in air is incident at grazing angle (incident angle $={90}^o$) on a long rectangular slab of a transparent medium of thickness $t=1.0m$. The point of incidence is the origin A(0, 0). The medium has a variable index of refraction n(y) given by
$n\left(y\right)=[k{y}^{3/2}+1{]}^{1/2}$
where $k=1.0(m{)}^{-3/2}$
a. Obtain a relation between the slope of the trajectory of ray at point B(x, y) in the medium and the incident angle at the 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 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.

If i is the angle of incidence at B(x, y), then slope of trajectory at B,

$dy/dx=tan\theta =tan({90}^o-i)=coti$ (i)

b.

From Snell's law, $nsini=constantC$.

From Snell's law at A(0, 0),

$nsini=1\times sin{90}^o=1\Rightarrow nsini=1$

$sini=1/n$ or $i=si{n}^{-1}\left(1/n\right)$

$\therefore coti=\frac{cosi}{sini}=\frac{\sqrt{(101/n{)}^2}}{\left(1/n\right)}=\sqrt{n^2-1}$

From Eq. (i),

or $n^{2}si{n}^{2}i$ or $n^2\frac{1}{1+co{t}^{2}i}=1$

or $\frac{n^2}{1+(dy/dx{)}^2}=1$

Given $n=[k{y}^{3/2}+1{]}^{1/2}\Rightarrow n^2=k{y}^{3/2}+1$

$\therefore \frac{k{y}^{3/2}+1}{1+(dy/dx{)}^2}=1\Rightarrow k{y}^{3/2}+1=1+{\left(\frac{dy}{dx}\right)}^2$

or $(dy/dx{)}^2=k{y}^{3/2}\Rightarrow dy/dx-k^{1/2}{y}^{3/4}$

or $dy/y^{3/4}=k^{1/2}dx$.

Integrating, we get $4y^{1/4}=k^{1/2}x+C$

where c is constant of integration.

At $x=0,y=0\Rightarrow C=0$

$\therefore 4y^{1/4}=k^{1/2}x$

As $k=1.0$ (given)

$\therefore y^{1/4}=\left(1/4\right)x$ (iii)

This is the required equation of trajectory.

c.

At $y=1.0m$, Eq. (ii) given $x=4m$.

$\therefore B(x_1,y_1)=P(4,1)$

d.

The path of the ray subsequently will be the grazing angle of emergence since , $nsine=1$ or $1sine=1\Rightarrow e=90$.