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A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40^(@). What is the refractive index of the materal of the prism? The refracting angle of the prism is 60^(@). If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of parallel beam of light. |
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Answer» Solution :Here, `angleA = 60^(@)` and `angleD_(m) = 40^(@)` `therefore` Refractive index of glass prism `n_(g) =(sin(A+D_(m))/2)/(sinA/2) =(sin(60 + 40)/2)^(@)/(sin(60/2)^(@)) = (sin 50^(@))/(sin 30^(@)) = 0.7660/0.5000 = 1.53` When prism is placed in water `(n_(W) = 1.33)`, then `n_(GW) = 1.53/1.33 = 1.15` `RARR sin(30^(@) + D_(m)^(.)/2) = 1.15 xx 0.5000 = 0.5750` or `(30 + D_(m)^(.)/2) = sin^(-1) (0.5750) = 35.1^(@)` or `D_(m)^(.) = 2 xx 5.1 = 10.2^(@) = 10^(@)` |
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