InterviewSolution
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InterviewSolution
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A ray of light, incident on an equilateral glass prism (mu_(g) = sqrt(3)) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray. |
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Answer» Solution :As the REFRACTED light MOVES parallel to the base line of the prism inside it, the prism is set in minimum DEVIATION condition. `angleA = 60^(@)`for an equilateral prism and REFRACTIVE index of glass `n_(g) = sqrt(3)` `therefore sqrt(3) =((sini 60^(@) + D_(m))/2)/(sin 60^(@)/2) =(sin(60^(@) + D_(m))/2)/(1/2)` `rArr (sin(60^(@) + D_(m))/2) = sqrt(3)/2 rArr (60^(@) + D_(m))/2 = sin^(-1) (sqrt(3)/2) = 60^(@) rArr D_(m) = 60^(@)` In minimum deviation condition angle of INCIDENCE `i=(A+D_(m))/2 =(60^(@) + 60^(@))/2 = 60^(@)` |
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