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A convec lens of focal length 0.24 m and of refractive index 1.5 is completelyimmeresed in water of refractive index 1.33, Find the change in the focal length of the lens. |
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Answer» SOLUTION :Given : - `n_g=1.5 , n_w =1.33` f=0.24 m When the lens is in air , `1/(f_g)=(n_g-1)[1/R_1-1/R_2]` `1/0.24=[1.5-1][1/R_1-1/R_2]` ….(1) When the lens is in water , `1/(f_(gw))=[n_g/n_w-1][1/R_1- 1/R_2] =[1.5/1.33-1] [1/R_1-1/R_2]` `1/(f_(gw))=[1.1278-1][1/R_1 - 1/R_2]` ...(2) DIVIDING eqn. (1) by eqn. (2)  `(f_(gw))/f_g=0.5/0.1278` `(f_(gw))/0.24=0.5/0.1278` `f_(gw)` = 0.94 m CHANGE in focal length = 0.94-0.24=0.70 m |
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