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.

Explanation

(a) For the image to form at the same place as that of object , light rays must fall normally to the silvered surface so that they are turned right back and rays traverse their path back to object .

Radius of curvature of convex surface which is silvered is 20 cm . Hence , for the ray to fall normally on this surface , image distance in first refraction must be 20 cm . Applying refraction formula .
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
Refraction is from air to glass hence , `mu_(1) = 1` and `mu_(2) = 1.5`
`implies " " (1.5)/((-20)) - (1)/(mu) = (1.5 - 1)/((-60))`
`implies - (3)/(40) - (1)/(u) = - (1)/(120)`
`implies (1)/(u) = (1)/(120) - (3)/(40)`
`implies " " (1)/(u) = (1- 9)/(120)`
`implies " " u = -15 ` cm
Hence , object must be placed at a distance 15 cm from the lens .
(b) When water is filled on concave surface of lens then first refraction will be from air to water then second refraction from water to glass and this time image distance in second refraction should be equal to radius 20 cm .
Note that we can use the same refraction formula for refraction at plane surface also by putting R = infinity .
We have to use refraction formula twice .
Air to water
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
`((4)/(3))/(v.) - (1)/(mu) = ((4)/(3) - 1)/(oo)`
`(4)/(3v.) - (1)/(u) = 0 " " ... (i)`
Image distance v. in previous refraction is object distance for next refraction from water to glass .
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
`implies (1.5)/(-20) - (3)/(v.) = - (5)/(3 xx 600)`
`implies " " - (3)/(40) - (4)/(3v.) = - (1)/(360)" " .... (ii) `
Adding the equation (ii) with equation (i) we get the following :
`implies " " - (3)/(40) - (1)/(u) = - (1)/(360)`
`implies " " (1)/(u) = (1)/(360) - (3)/(40)`
`implies " " (1)/(u) = (1- 27)/(360)`
`implies u = - (360)/(26)` cm
`implies " " u = -13.85` cm
Since new location of object is at 13.85 cm from the lens , hence , displacement from its earlier position is `15 - 13.85 = 1.15` cm .
Since new distance is less than that in earlier case hence object is to be moved by a distance 1.15 cm towards the lens .