question_answer

An object is placed at \[f/2\] away from first focus of a convex lens where \[f\] is the focal length of the lens. Its image is formed at a distance \[3f/2\] in a slab of refractive index \[3/2\], from the face of the slab facing the lens. Find the distance of this face of the slab from the second focus of the

A) \[f/2\]       

B)                    \[3f/2\]

C) \[2f\]                                    

D) \[f\]

Correct Answer: D

Solution :

\[u=-\left( f+\frac{f}{2} \right)=-\frac{3f}{2}\] Distance of final image from slab\[=\frac{3f}{2}\] Distance of image formed due to convex lens from face of the slab facing the lens. \[=\frac{(3/2)f}{\mu }=f\] Let          \[{{F}_{2}}S=x\] \[v=2f+x\]                 From \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\frac{1}{(x+2f)}+\frac{1}{(-3f/2)}=\frac{1}{f}\]or\[x=f\]