Question

Two thin convex lenses of focal length $f_1$ and $f_2$ are separated by a horizontal distance $d$ (where $d<f_1,d<f_2)$ and their centers are displaced by a vertical separation $△$ as shown in the figure.
Taking the origin of coordinates $O$ at the center of the first 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. $x=\frac{f_1(f_2+d)}{f_1+f_2-d},y=\frac{\Delta }{f_1+f_2}$
• 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. $x=\Delta ,y=\frac{\Delta (f_1-d)}{f_1+f_2-d}$

Answer 1

Answer: (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 image $I^{\prime }$ of parallel rays formed by lens $1$ will act as a virtual object.

Applying lens formula for lens $2$,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}-\frac{1}{f_1-d}=\frac{1}{f_2}$

$\Rightarrow v=\frac{f_2(f_1-d)}{f_2+f_1-d}$

The horizontal distance of the image $I$ from $O$ is

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

$=\frac{d{f}_2+d{f}_1-d^2+f_2{f}_1-d{f}_2}{f_2+f_1-d}$

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

To find the y-coordinate, we use magnification formula for lens $2$,

$m=\frac{v}{u}=\frac{\frac{f_2(f_1-d)}{f_1+f_2-d}}{f_1-d}=\frac{f_2}{f_1+f_2-d}$. Also

$m=\frac{h^2}{△}\Rightarrow h_2=\frac{△\times f_2}{f_1+f_2-d}$

Therefore, the y-coordinate,

$y=△-h_2$

$=△-\frac{△f_2}{f_1+f_2-d}$

$=\frac{△f_1+△f_2-△d-△f_2}{f_1+f_2-d}==\frac{△(f_1-d)}{f_1+f_2-d}$.