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A concave lens has the same radii of curvature for both sides and has a refractive index 1.6 in air. In the second case, it is immersed in a liquid of refractive index 1.4. Calculate the ratio of the focal lengths of the lens in two cases. |
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Answer» Solution :When the LENS PREPARED from a material of REFRACTIVE index `n_L = 1.6` is placed in air, then its focal length `f_("air")` is GIVEN by: `1/(f_("air") =(n_(L)-1) (1/R_(1)-1/R_(2))) = (1.6-1) (1/R_(1)-1/R_(2))` When the lens is immersed in a medium of refractive index `n_m = 1.4`, then its new focal length `f_(m)`is given by: `1/f_(m) = (n_(L)/n_(m)-1)(1/R_(1) - 1/R_(2)) = (1.6/1.4-1)(1/R_(1)-1/R_(2))` `RARR f_(m)/f_("air") =((1.6-1) xx 1.4)/(1.6 - 1.4) = (0.6 xx 1.4)/2 = 4.2` or `f_(m)=4.2 f_("air")` |
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