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Lens Formula

Two thin conv...

Question

Two thin convex lenses of focal length $f_{1} a n d f_{2}$ are separated by a horizontal distance d (where $d < f_{1}, d < f_{2}$ ) and their centers are displaced by a vertical separation $Δ$ as shown in 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 = Δ

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B

x = \frac{f_{1} (f_{2} + d)}{f_{1} + f_{2} - d}, y = \frac{Δ}{f_{1} + f_{2}}

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C

x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = \frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d}

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D

x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = \frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d}

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Solution

The correct option is C $x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = \frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d}$
patiently use lens formula for individual lens one by one and get the final position of the image. The total distance would be d+v(final) and which will come as $\frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}$ .
for finding y, use magnification formula and get $\frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d}$
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