In figure , L is half part of an equiconvex glass lens `(mu=1.5)` whose surfaces have radius of curvature R =40cm and its right surface is silvered . Normal to its principal axis a plane mirror M is placed on right of the lens . Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror , calculate distance'a' in cm between lens and object.

Explanation

Distance of image of object O from plane mirror = a+b . Since , there is no parallax between the image formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point . Distance of image = (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 , 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.
`therefore (a+ 2b) = 2a implies b = (a)/(2)`
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 figure.

Focal length of convex lens `f_(1)` is given by `(1)/(f_(1)) = (mu-1)((1)/(R_(1)) - (1)/(R_(2))) implies f_(1) = 40"cm"`
For concave mirror focal length , `f_(m) = (R)/(2)= -20"cm"`
The combination L behaves like a mirror whose equivalent focal length F is given by
`(1)/(F) = (1)/(f_(m)) - (2)/(f_(1)) implies F= -10"cm"`
Hence , for the combination u=-a , v=+2a , F=-10cm
Using mirror formula , `(1)/(v) + (1)/(u) = (1)/(F) implies a = 5"cm"`