The convex surface of a thin concave-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image is formed at the same place ? (b) If the concave part is filled with water (mu = 4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.

Answer

(a) `15cm` form the lens on the axis (b) `1.14cm` towards the lens.

Explanation

First image should from at centre pf curvature fo mirror after refraction by convave surfaces.
`(1.5)/(-20) - (1)/(u) = (1.5-1)/(-60), (1)/(u) = (1)/(120) - (1.5)/(20) = (1-9)/(120), u = -(120)/(8) = -15cm.`
(b) After refraction by water lens image should form at `15cm` (Answer of part a)
`(1)/(f_("water")) = (1)/(-15) - (1)/(u) = ((4)/(3) -1) ((1)/(oo) - (1)/(-60))`
`rArr u = -13.86 cm`
`rArr` So, distance through which the pin should be moved `= (15-13.86) = 1.14cm` towards lens.