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A double convex lens, made from a material of refractive index mu_1,is immersed in a liquid of refractive index mu_2, where mu_2 gt mu_1 . What change,if any, would occur in the nature of the lens? |
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Answer» Solution :Focal LENGTH of lens (REFRACTIVE INDEX `mu_1` ) in a liquid of refractive index `mu_2`is Formula:` f_1=(mu_1 - 1)/(mu_1/mu_2) xx f_a` Given ` mu_2 GT mu_1 , i.e., mu_1/mu_2 lt 1` so ` f_1 = (mu_1 - 1)/(1 - mu_1/mu_2) f_a` So the focal length of lens in liquid will be of opposite sign of the focal length of lens in air, i.e., nature of lens will change. Hence, lens would now behave like a DIVERGING (concave) lens. |
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