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A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33. (a) Will it behave as a converging or a diverging lens in the two cases ? (b) How will its focal length change in the two media ? |
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Answer» Solution :Let FOCAL LENGTH of given convex lens in air `f_("air") =f` (say) , refractive index of glass `n_g = 1.5` (a) (i) When the lens is dipped in a medium of refractive index `n_m = 1.65` , the lens will start BEHAVING as a diverging lens because `n_m gtn_g`. (ii) However, when the lens is dipped in a medium of refractive index n.m = 1.33, the lens will continue to behave as a converging lens because `n_(m)^(.) lt n_(g)`. (b) We know that focal length of a lens in a medium `f_(m)`is given by: `f_(m) = (n_(m)(n_(g)-1))/(n_(g)-n_(m))f`, where f= focal length of the lens in air. (i) When glass lens (`n_(g) = 1.5`) is dipped in 1ST medium (`n_m = 1.65`), then `f_(m) = (1.65(1.5-1))/(1.5-1.65).f = (1.65 xx 0.5)/(-0.15).f = -5.5 f` (ii) When glass lens (`n_(g)=1.5`) is dipped in 2ND medium `(n_(m)^(.) = 1.33)`then `f_(m)^(.) = (1.33(1.5-1))/(1.5 -1.33).f = (1.33 xx 0.5)/(0.17)f. = 3.91 f` |
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