Q. 47

# The convex surface of a thin concavo-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 (18-E13). (a) Where should the 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 (μ =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) Let F be the focal length of the given concavo-convex lens.

Then,

--------------- (i)

The value of can be obtained by using lens makers’ formula:

Substituting these values in equation (i):

F=

For the image to be informed at the same point as the object

u = 2F = 2 x 7.5 = 15 cm.

Therefore, the pin should be placed at a distance of 15 cm from the lens.

(b) If the concave part is filled with water;

If is the focal length of lens in water, the focal length F’ of this combination is given by:

----------------------- (ii)

The value of is calculated by Lens makers’ formula:

⇒

⇒

Substituting these values in equation (ii):

u’ = 2F= 2× = cm

Displacement of pin = u-u’ = (towards the lens).

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