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(a) A point object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices nx and n_(2)(n_(2) gt n_(1)) Draw the ray diagram and deduce the relation between the object distance (w), image distance (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium. (b) A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, find its new focal length. |
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Answer» Solution :(a) N/A (b)Focal LENGTH of lens in air f= + 20 cm, refractive index of glass ng = 1.6 and refractive index of liquid `n_(L) = 1.3` If radii of CURVATURE of two surfaces be `R_1` and `R_2`, respectively, then in air: `1/f =(n_(g)-1)(1/R_(1)-1/R_(2))` i.e. `1/20 = (1.6-1) (1/R_(1) -1/R_(2))` .........(i) and on immersing the lens in given liquid, new focal length f.will be given by: `1/f^(.) =(n_(g)/n_(1)-1) (1/R_(1)-1/R_(2)) = (1.6/1.3-1) (1/R_(1)-1/R_(2))` ...........(ii) Dividing (i) by (ii), we get `f^(.)/20 =(0.6 xx 1.3)/0.3 rArr f^(.) = 52 cm` |
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