Question

The radius of curvature of the curved surfaces of an equiconvex lens is 32 cm and its refractive index is `mu = 1.5`. One of its side is silvered and placed 14 c away from an object as shown in figure. At what distance x should a second convex lens of focal length 24 cm be placed so that the image coincides with the object.

Answer 1

Equivalent focal length of silvered lens
`(1)/(F)=(1)/(f_l)+(1)/(f_m)+(1)/(f_l)=(2)/(f_l)+(1)/(f_m)`
`=2((3)/(2)-1)((1)/(32)-(1)/(-32))+(1)/((32)/(2))`
`=(1)/(16)+(1)/(16)=(1)/(8)impliesF=8cm`
The image by lens should be formed at centre of curvature of mirror.
`(1)/(-(16-x))-(1)/(-(14-x))=(1)/(24)`
`(2)/((16-x)(14-x))=(1)/(24)implies48=224-30x+x^2`
`x^2-30x+176=0`
`x^2-22x-8x+176=0`
`x(x-22)-8(x-22)=0`
`(x-8)(x-22)=0`
`x=8` or `22cm` is not possible
`x=18cm`