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Asked by sarveshvibrantacademy | 16 May, 2019, 10:53: AM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/0ab11ea075c5324caf6fa349a6f19e6f5cdd7f6e643f00.57245846ri.png)
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) =
..........................(2)
![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](https://images.topperlearning.com/topper/tinymce/cache/0a154393923e26091d9884a0a2f0c8a1.png)
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 Thiyagarajan K | 16 May, 2019, 21:27: PM
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