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

(1) In situation where refractive index changes only in one direction, medium can be considered as a collection of large number of the thin layers. Let refractive index be a function of y, that is n=f(y) then medium can be considered as to made up of a large number of thin slabs placed parallel to x axis and normal at any surface is parallel to x axis and normal at any surface is parallel to y axis.
(2) Although the refractive index is not constant, we cannot apply Snell.s laws here, but still we can divide the whole medium into thin layers as shown in Fig. 34-22. Within a layer, refractive index is constant. So here we have a case of many parallel slabs. We can apply Snell.s law here.
`n_(1) sin theta_(1)=n_(2)sintheta_(2)=n_(3)sintheta_(3)=.....`
`nsintheta`=constant
Calculations
(a) From Fig. 34-22b,
`i+theta=90^(@)`
Slope of the tangent `=tantheta`
`=tan(90-i)=coti`
(b) We know that at any point in the medium with variable refractive index n sin i constant, where n= refractive index at any point y and i=angle of incidence at any point y. Therefore,
`nsini=1sin90^(@)`
`sini=(1)/(coseci)=(1)/(sqrt(1+cot^(2)i))`
Substituting the values `(ky^(3//2)+1)^(1//2)`, we get
`(1)/(sqrt(1+m^(2))=1`
`ky^(3//2)+1=1+m^(2)`
Slope `m=(dy)/(dx)=sqrt(k)y^(3//4)`
`y^(-3//4)dy=sqrt(k)dx`.
On integrating both sides, we get
`int_(0)^(y)y^(-3//4)dy=int_(0)^(x)sqrt(k)dx`
`4y^(1//4)=x` (since `k=1`)
`y=(x^(4))/(256)`
(c ) For `y=1m`, `x=4m`
(d) Let `alpha_(e )` be the angle of emergence in the air, then by Snell.s law ,
`n_(e )sinalpha_(e )=n_(1)sinalpha_(1)=1` (refractive index of air `n_(e)=1`)
This gives
`sinalpha_(e )=1impliesalpha_(e )=90^(@)`
Thus, it emerges parallel to the incident ray , this could have been interpreted by Snell.s law too.