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Asked by Sunita | 24 Sep, 2018, 09:20: PM
answered-by-expert Expert Answer
when the light ray incident on glass Slab-1 with incidence angle i,
we have as per laws of refraction, begin mathsize 12px style fraction numerator sin space i over denominator sin space r subscript 1 end fraction space equals space mu subscript 1 space....................... left parenthesis 1 right parenthesis end style
where r1 is the angle of refraction in slab-1
when the refracted ligh ray in slab-1 travels to slab2, then we have

begin mathsize 12px style fraction numerator sin space r subscript 1 over denominator sin space r subscript 2 end fraction space equals space mu subscript 2 over mu subscript 1 space....................... left parenthesis 2 right parenthesis end style
where r2 is the angle of refraction in slab-2

when the refracted light ray emerging out of slab-2, then we have
begin mathsize 12px style fraction numerator sin space r subscript 3 over denominator sin space r subscript 2 end fraction space equals space mu subscript 2 space space....................... left parenthesis 3 right parenthesis end style
θ = 90 - r3 ................................(4)
 
from eqn.(3), eqn.(2) and eqn.(1), we have,  begin mathsize 12px style sin space r subscript 3 space equals space mu subscript 2 space sin space r subscript 2 space equals space mu subscript 2 space sin left parenthesis r subscript 1 right parenthesis cross times mu subscript 1 over mu subscript 2 space equals space sin space i end style ...........(5)
hence angle r3 = angle i  or θ = 90 - i
Answered by Thiyagarajan K | 25 Sep, 2018, 12:34: PM

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