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

Let `theta` be the angle of incidence at point `P(x,y)` then according to Snell.s law we can write the following
`upsilon sin theta=c`
Here `c` is a constant.
At point `O` we know that `theta=90^(@)`
and `upsilon=(a^(2)y^(3//2)+1)^(1//2)=1(therefore y=0)`
Hence we get `c=1.`
Hence `upsilon sin theta=1`
`implies " " sin theta =(1)/(upsilon)` " ".......(i)
Here angle `theta` is measured with normal which is parallel to `Y=`axis. Hence angle the light ray makes with the horizontal is `90^(@)-theta`. Hence, slope at the point `P` can be written as follows:
`(dy)/(dx)=tan(90^(@)-theta)`
`implies" "(dy)/(dx)=cot theta`
From equation `(i)` we can find the value of cot `theta` which is equal to `sqrt(upsilon^(2)-1)`, hence we can write the following:
`implies " "(dy)/(dx)=sqrt(upsilon^(2)-1)`
On substituting given expression of `upsilon` in above equation we get the following:
`implies " " (dy)/(dx)=sqrt(a^(2)y^(3//2)+1-1)`
`implies " "(dy)/(dx)=ay^(3//4)`
`implies " " y^(-3//4)`
On integrating both the sides we get the following :
`int y^(-3//4)dy=a int dx`
`implies" " 4y^(1//4)=ax+c_(1)`
At `x=0,y=0` and hence `c_(1)=0,` so equation can be written as follows:
`4y(1//4)=ax`