The radii of curvature of each surface of a convex lens is `20 cm` and the refractive index of the material of the lens is `3//2` (i) Calculate its focal length (ii) If this is cut along the plane `AB`. What will be formed ? (iii) What happens if the lens is cut along `CD` ?

Answer

`20 cm ; 40 cm ; 20 cm`

Explanation

(i) `(1)/(f) = (mu - 1) ((1)/(R_1) - (1)/(R_2))`
=`((3)/(2) - 1) ((1)/(20) + (1)/(20)) = (1)/(20)` ,
`f = -20 cm`
(ii) When lens is cut along the plane `AB`, Fig., focal length of each half is given by
`(1)/(f) = (mu - 1) ((1)/(R_1) - (1)/(oo)) = ((3)/(2) - 1) xx (1)/(20) = (1)/(40)`
`f_1 = 40 cm`
(iii) When lens is cut along `CD`, Fig., each half continues to have original focal length `f = 20 cm`. However, intensity of image will reduce.
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