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Asked by sarveshvibrantacademy 16th May 2019, 10:53 AM
Answered by Expert
As shown in figure let a light ray enters at P and makes angle of incidence θ with y-axis. 
Let the refractive index of the medium changes as μx = 1+x2 . Let the ray makes angle of incidence θx at Q in the medium.
If the given medium is in the form of rectangular slab, we have,   1×sinθ  = μx × sinθx   or  sinθx = sinθ / μx  ..................(1)
hence we have,     tanθx = (dy/dx) = begin mathsize 12px style fraction numerator begin display style bevelled fraction numerator sin theta over denominator mu subscript x end fraction end style over denominator square root of 1 minus space space begin display style fraction numerator sin squared theta over denominator mu subscript x superscript 2 end fraction end style space end root end fraction space equals space fraction numerator sin theta over denominator square root of mu subscript x superscript 2 minus sin squared theta end root end fraction end style ..........................(2)
if the incident light ray is nearly parallel to y-axis, then agle of incidence θ = π/2 . 
At x = 1, refractive index μx = 1+x2 = 1+12 = 2 .  
By substituting values in eqn.(2), we get (dy/dx) at x=1 for the light ray entering almost parallel to y-axis as  (dy/dx)x=1 = 1/√3
Hence unit vector of the light ray at the point Q = (√3/2) i + (1/2) j , where i and j are unit vectors along x and y directions
Answered by Expert 16th May 2019, 9:27 PM
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