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.

Answer

A, B, C, D

Explanation

(a) SLOPE AT P
To find the slope at B, we draw a tangent to the trajectory at B. The trajectory is such that as the ray passes through the rectangular transparent medium, the ray continuously deviates towards the normal. The tangent at B makes an angle `theta` with the x-axis. Therefore, the slope at point B is `tantheta=(dy)/(dx)....(i)
i is the angle of incidence at B then according to `triangle`BQM
`i+theta+(pi)/(2)=pi`....(ii)
Substituting the value of `theta` form (ii) in (i)
`((pi)/(2)-i)=(dy)/(dx)implies(dy)/(dx)=coti` ...(iii)
(b) EQUATION OF TRAJECTORY
According to snell's law, when light propagates through a series of parallel layers of different media, then `nsini=constant`
Let us consider the rectangular state to be made up of parallel layers such that as we move in the +Y direction, the refractive index increases as given by the relationship
`n(y)=[ky^((3)/(2))+1]^((1)/(2))` ,,,(iv)
Applying Snell's law at B, we get `nsini=const.=1(from above equation)`
`:'``n=(1)/(sini)=coseci=sqrt(1+cot^(2)i)=sqrt(1+((dy)/(dx))^(2))`,
`sqrt(ky^((3)/(2))+1)=sqrt(1+((dy)/(dx))^(2))` from (iv) `implies``=(dy)/(dx)=[ky^((3)/(2))]^((1)/(2))implies(dy)/(y^((3)/(4)))=k^((1)/(2))dx=dx` (`:"`k=1)`
`impliesint(dy)/(y^((3)/(4)))=intdx`
`implies` y^((1)/(4))=x+C` where C is an integration constant. But at `x=0`,`y=0`
`:'` `C=0` `:'` `4y((1)/(4))=x=impliesy=((x)/(4))^(4)`
(c) CO-ORDINATES `(x_1,y_1)` OF THE POINT P
At P,y=1m `:'``x=4y_((1)/(4))=4`
The coordinates of P are `(4m,1m)`
(d) The refractive index P
`n_p=[ky^((3)/(2))]^((1)/(2))=[1(1)^((3)/(2))+1]^((1)/(2))=sqrt2`
if `i_p` is angle of incidence at P then accordint to Snell's law,
`n_(p)sini_(p)=1impliessini_p=(1)/(sqrt2)`
Also by Snell's law, `n_(air)sinr_p=n_(p)sini_(p)`
`1sinr_(p)=sqrt2xx(1)/(sqrt2)impliessinr_(p)=1impliesr_(p)=(pi)/(2)` `implies` After emerging from the rectangular glass slab, the light tay becomes parallel to slab length.