A ray of light travelling in air is incident at grazing angle (incident angle=`90^(@))` on a long rectangular slab of a trasnparent 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)`,wherek=`1.0m^(-3//2)`.the refrative 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

(i)`tantheta=(dy)/(dt) =coti;(ii)1,(iii)y=k^(2)(x//4)^(4); (iv)4.0,1`;(v) it will become parallel to x-axis.

Explanation

(i)slope =`(dy)/(dx)=tantheta rArr (dy)/(dx)=tan(90-i)`
`(dy)/(dx)=cot i ……….(1)`
(ii) Applying snell's law between point of incidence and at point B.
`n_(y).sini=1xxsin90^(@)`
`n_(y).sin i=1`
(iii)`[ky^(3//2)+1]^(1//2)(1)/(cosec i)=1 rArr [ky^(3//2)+1]^(1//2)(1)/(sqrt(1+cot^(2)i))=1`
`[ky^(3//2)+1]^(1//2)=sqrt((1+cot^(2))i)`.....(2)
From (1)&(2)
`[ky^(3//2)+1]=1+((dy)/(dx))^(2) rArr ky^(3//2)=((dy)/(dx))^(2)rArr (dy)/(dx)=sqrt(k)y^(3//4)`
`underset(0)overset(4)inty^(-(3)/(4))=sqrt(k)underset(0)overset(X)intdx rArr[y^[-(3)/(4)+1)/(-(3)/(4)+1)]_(0)^(y)=sqrt(k)[X]_(0)^(X)rArr4y^(1//4)=sqrt(k)X`
`y=k^(2)((x)/(4^(4)))`
(iv)`1=k^(2)((x)/(4^(4))) rArr 1=(1)^(2).((x)/(4^(4)))`
`x=4`
(iv)Ray will emage parallel to incident ray means parallel to x-axis.