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Asked by sarveshvibrantacademy | 27 Apr, 2019, 11:48: AM
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Figure shows the incident ray travelling along the vector direction ( -i -2j )  in a medium that has absolute refrative index μ = 2
and getting refracted in another medium that has absolute refrative index μ = √5/2
 
Angle θ made by incident ray with x-axis is given by,  θ =  tan-1(2) , hence sin i  = sin (90-θ) = cos θ = 1 / √5 .
 
As per law of refraction = sin i / sin r = (√5 /2) / 2  = √5 /4
 
hence sin r = ( 1/ √5 ) / (√5 /4 )  = 4/5  or tan (r) = 3/4 
 
hence slope of the line that represents the refracted ray = 4/3.
 
Hence a vector along the refracted ray is of the form   (-3i -4j ) or unit vector  (-3i -4j)/5
Answered by Thiyagarajan K | 27 Apr, 2019, 19:40: PM

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