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, B, C

Explanation

(i) KEY CONCEPT: The given silvered concavo-convex lens behaves leke a mirror whise focal length cabe
calculated by the formula `(1)/(f)=(2)/(f_(1))+(1)/(f_(2))`
`f_(1)=`focal length of concave surface.
`f_(2)=` focal length of concave mirror.
`:'``(1)/(f)=(2)/(-30)+(1)/(-10)=-(4)/(30)`
`:'` `f=-7.5cm`
Using mirror formula
`(1)/(f)=(1)/(v)+(1)/(u)implies(1)/(-7.5)=(1)/(-x)+(1)/(-x)`
`x=15cm`

(ii) Let the object distance be u. When water is poured over the concave surface the apparent object distance will be v then
`-(mu_(1))/(u)+(mu_(2))/(v)=(mu_(2)-mu_(1))/(R)`
For flat suface `R=infty`
`:'``-(mu_(1))/(u)+(mu_(2))/(v)=0`implies`v=u(mu_(2))/mu_(1)=uxx_1^2mu=uxx(4)/(3)`
Since the ray enters the lens from water into glass
`(-mu_(w))/(u)+(mu_(g))/(v)=(mu_(g)-mu_(w))/(-60)impliesu=-13.85`cm` `:'` Downward shift`=15-13.85=1.15cm.`