A ray of light travelling in air is incident at grazing angle (incident angle=`90^(@))` on a long rectangular slab of a transparent medium of thickness `t=1.0` (see figure). The point of incidence is the origin `A(O,O)` .The medium has a variable index of refraction n(y) given by :`n(y)=[ky^(3//2)+1]^(1//2)` ,where k=`1.0m^(-3//2)`.the refractive index of air is 1.0`

(i) 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
(ii) obtain an equation for the trajectory `y(x)`of the ray in the medium.
(ii) Determine the coordinates (`x_(1),y_(1))` of the point `P`.where the ray the ray intersects upper surface of the slab -air boundary.
Indicate the path of the ray subsequently.

Explanation

a. If i is the angle of incidence at `B(x,y)`, then slope of trajectory at B,
`dy//dx=tantheta=tan(90^@-i)=cot i` (i)
b. From Snell's law, `n sini=constant C`.
From Snell's law at `A(0,0)` ,
`n sin i=1xxsin90^@=1rArrnsini=1`
`sini=1//n` or `i=sin^(-1)(1//n)`
`:. cot i=(cos i)/(sin i)=sqrt((1-1//n)^(2))/(1//n)=sqrt(n^(2)-1)`
From Equ. (i),
or `n^(2)sin^(2)i` or `n^(2)(1)/(1+cot^(2)i)=1`
or `n^(2)/(1+(dy//dx)^(2))=1`
Given `n=[ky^(3//2)+1]^(1//2)rArr n^(2)=ky^(3//2)+1`
`:. (ky^(3//2)+1)/(1+(dy//dx)^(2))=1rArr ky^(3//2)+1=1+((dy)/(dx))^(2)`
or `(dy//dx)^(2)=ky^(3//2)rArr 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 rArr C=0`.
`:.4y^(1//4)=k^(1//2)x`
As `k=1.0` (given)
`:. y^(1//4)(1//4)x` (iii)
This is the required equation of trajectory.
c. At y=1.0 m, Eq. (ii) gives `x=4m`.
`:. 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 `1 sine=1rArr e=90`.