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Common Rays in Thin Lenses

Two thin conv...

Question

Two thin convex lenses of focal lengths $f_{1}$ and $f_{2}$ are separated by a horizontal distance $d (d < f_{1}$ and $d < f_{2})$ and their centers are displaced by a vertical separation as shown in figure. A parallel beam of rays coming from left. Take the origin of coordinates $O$ at the center of first lens.
Find the x-coordinates in the same problem from the focal point of this lens system.

A

\frac{d (f_{1} - d) + f_{1} f_{2}}{(f_{1} + f_{2} - d)}

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B

\frac{f_{1} f_{2}}{(f_{1} + f_{2} - d)}

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C

\frac{d (f_{1} - d)}{(f_{1} + f_{2} - d)}

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D

\frac{2 d (f_{1} + d) - f_{1} f_{2}}{(f_{1} + f_{2} - d)}

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Solution

The correct option is A $\frac{d (f_{1} - d) + f_{1} f_{2}}{(f_{1} + f_{2} - d)}$
The parallel rays will be focussed at the focal point of the first lens. The first image lies at $I$ , at a distance $f_{1}$ from the origin. This image $I_{1}$ will act as an object for refraction through the second lens. The object distance for the second lens, $u = (f_{1} - d)$ .
From lens equation, $\frac{1}{v} - \frac{1}{+ (f_{1} - d)} = \frac{1}{f_{2}}$
$v = \frac{f_{2} (f_{1} - d)}{(f_{1} + f_{2} - d)}$
Hence, the x-coordinate of final image $I_{2}$ is
$x = d + u = \frac{d + f_{2} (f_{1} - d)}{(f_{1} + f_{2} - d)} = \frac{d (f_{1} - d) + f_{1} f_{2}}{(f_{1} + f_{2} - d)}$

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