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An equiconvex lens of refractive index n, focal length/and radius of curvature R is immersed in a liquid medium of refractive index n_m. For (?) n_m gt n, and (ii) n_m lt n, draw the ray diagrams in the two cases when a beam of light coming parallel to the principal axis is incident on the lens. Also find the focal length of the lens in terms of the original focal length and the refractive indices of the lens and that of the medium. |
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Answer» Solution :Ray diagrams have been shown in FIG. 9.62 (i) and (ii) RESPECTIVELY. For an equiconvex lens `|R_(1)| = |R_(2)| =R` and sign of `R_(1)`is positive but that of `R_2` is negative. Hence, in air MEDIUM, the focal length of lens is given as :  `1/f = (n-1) [1/(+R) - 1/(-R)]` `rArr 1/f = (n-1) 2/R`..........(i) When the lens is immersed in a liquid medium, its focal length changes to `f_(m)` where `1/f_(m) =(n/n_(m)-1) 2/R` dividing (i) by (ii), we GET `f_(m)/f = (n-1)/(n/n_(m)-1) = ((n-1)n_(m))/(n-n_(m)) rArr f_(m) ((n-1)n_(m))/(n-n_(m)).f` |
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