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Convex lens is made of glass of refractive index `1.5` If the radius of curvature of each of the two surfaces is `20 cm` find the ratio of the powers of the lens, when placed in air to its power, when immersed in a liquid of refractive index `1.25`. |
Answer» Correct Answer - `5//2` Here, `mu_(g) = 1.5, R_(1) = 20 cm = 0.2 m` `R_(2) = - 20 cm = - 0.2m` `P_(1) = (1)/(f_(1)) = ((mu_(g))/(mu_(a)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= (1.5 - 1)((1)/(0.2) + (1)/(0.2)) = 5 D` When the same lens is placed in liquid, `P_(2) = (1)/(f_(2)) = ((mu_(g))/(mu_(l)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `= ((1.5)/(1.25) - 1)((1)/(0.2) + (1)/(0.2)) = (0.25)/(1.25) xx (2)/(0.2) = 2 D` `(P_(1))/(P_(2)) = (5)/(2)` |
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