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The radii of curvatore of two surfaces of a convex lens is 0.2 m and 0.22 m. Find the the forcal length of the lens if refractive index of the material of lens is 1.5. Also find the change in focal length, if it is immersed in water of refractive index 1.33. |
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Answer» Solution :GIVEM: ` R_(1) + 0.2m ` ` R_(2) = - 0.22 m ` , ` n = 1.5 , n_(w) = 1.33 ` w.k.t ` 1/t = (n_(2) - 1) ((1)/(R_(1)) - (1)/(R_(2))) ` i.e., ` 1/f = (1.5 -1) ((1)/(0.2) + (1)/(0.22)) ` i.e., ` 1/f = 0.5 xx (5 + 4.550 = 4.78` Hence `f = (1)/(4.78) = 0.209 m" but " (f_(w))/(f_(a)) = (n_(r) -1)/(t^(n_(G)) -1) ` i.e., ` (f_(w))/(f_(a)) = (1.5 -1)/((1.33)/(1.33)-1) = 3.91 ` i.e., ` f_(w) = 3.91 f_(a) = 3.91 xx 0.209 = 0.817 m ` FOCAL length of LENS in water = 0.817 m |
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