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A ray of light suffers minimum deviation, while passing through a prism of refractive index `1.5` and refracting angle `60^@`. Calculate the angle of deviation and angle in incidence. |
Answer» Correct Answer - `37.2^(@), 48.6^(@)` `mu = 1.5, A = 60^(@), delta_(m) = ?, i = ?` As `(sin (A + delta_(m))//2)/(sin A//2) = mu` `:. (sin (60^(@) + delta_(m))//2)/(sin 60^(@)//2) = 1.5` `sin (60^(@) + delta_(m))//2 = 1.5 sin 30^(@) = 0.75` `(60^(@) + delta_(m))/(2) = sin^(-1)(0.75) = 48.6^(@)` `delta_(m) = 2 xx 48.6 - 60^(@) = 37.2^(@)` `i = (A + delta_(m))/(2) = (60^(@) + 37.2)/(2) = 48.6^(@)` |
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