Sol. Since, distance fo object O from plane mirror M is (a + b), therefore, it forms a virtual image at a distance (a + b) behind itself or a distance 2 (a + b) from the object O.

Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point.

Hence, distance of image from lens L is 2 (a + b) – a = (a + 2b) behind lens.

Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, modulus of transverse magnification produced by the lens L is equal to 2.

Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a.

∴ (a + 2b) = 2a or b = a/2 … (1)

The silvered lens L may be assumed as a combination of an equi-convex lens and a concave mirror placed in contact with each other co-axially as shown in Fig.

For the lens,

R 1 = + r, R 2 = – r, µ = 1.5

∴ Its focal length f 1 is given by

= (µ – 1) or f 1 = 40 cm

For concave mirror, R = – r = – 40 cm

∴ Its focal length, f m = = – 20 cm

The combination L behaves like a mirror whose equivalent focal length F is given by

= – or F = – 10 cm

Hence, for the combination,

µ = – a, v = + 2a, F = – 10 cm

Using mirror formula, + =

a = 5 cm

Substituting a = 5 cm in equation (1),

b = 2.5 cm