Two thin convex lenses of focal length f1 and f2 are separated by a horizontal distance d (where d < f1 and d < f2) and their centres are displaced by a vertical separation $∆$ as shown. Taking the origin of coordinates O at the centre of first lens, 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:

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

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

$x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = 0$

$x = \frac{f_{1} f_{2}}{f_{1} + f_{2}}, y = Δ$

answer is C.

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Detailed Solution

Image I1 from first lens will act as virtual object for second lens. For second lens –
Question Image
$u = (f_{1} - d); f = + f_{2}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{f_{2}} + \frac{1}{(f_{1} - d)}$

   $v = \frac{f_{2} (f_{1} - d)}{f_{2} + f_{1} - d}$
x-coordinate $= d + \frac{f_{2} (f_{1} - d)}{f_{2} + f_{1} - d} = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}$
Linear magnification,    $m = \frac{v}{u} = \frac{f_{2}}{f_{1} + f_{2} - d}$
The second image will be formed at    $m Δ$
So, y-coordinate $= Δ - m Δ$
$= Δ (1 - \frac{f_{2}}{f_{1} + f_{2} - d})$

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