Question

Two thin convex lenses of length $f_1$ and$f_2$ are separated by a horizontal distance $d$ (where $d<f_1$ and $d<f_2$ ), and their centers are displaced by a vertical separation $\Delta$ as shown in the figure. Taking the origin of coordinates, O at the center of left lens, the $x$ and $y$ coordinates of the focal point of this lens system for a parallel beam of rays coming from the left are given by

• A. $x=\frac{f_1{f}_2}{f_1+f_2},y=\Delta$
• B. $x=\frac{f_1(f_2+d)}{f_1+f_2-d},y=\frac{{\Delta }^2}{f_1+f_2}$
• C. $x=\frac{f_1{f}_2+d(f_1-d)}{f_1+f_2-d},y=\frac{\Delta (f_1-d)}{f_1+f_2-d}$
• D. $x=\frac{f_1{f}_2+d(f_1-d)}{f_1+f_2-d},y=0$

Answer 1

The correct option is C $x=\frac{f_1{f}_2+d(f_1-d)}{f_1+f_2-d},y=\frac{\Delta (f_1-d)}{f_1+f_2-d}$

The light coming from infinity is refracted from first lens to form image at the focus of the lens, that is, at a distance of $f_1$ from the first lens.

This image acts as object for second lens at a distance of $f_1-d$ from it.

Hence from the lens equation,

$\frac{1}{v}=\frac{1}{f_2}+\frac{1}{u}$

$⟹v=\frac{f_2(f_1-d)}{f_1+f_2-d}$

Hence the x-coordinate of this final image formed $=d+v$

$⟹x=\frac{f_1{f}_2+d(f_1-d)}{f_1+f_2-d}$

The magnification caused by the second lens is:

$\frac{v}{u}$$=\frac{f_2}{f_1+f_2-d}=\frac{h_2}{h_1}=\frac{{\Delta }^{\prime }}{\Delta }$

$⟹y=\Delta -{\Delta }^{\prime }$$=\frac{\Delta (f_1-d)}{f_1+f_2-d}$