Two converging lenses have focal length f₁ and f₂ (f₁ >f₂). The optical axis of the two lenses coincide. This lens system is used to from an image of real object. It is observed that final magnification of the image does not depend on the distance x. Whole arrangement is shown in figure. Final magnification is :

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$- \frac{f_{1}}{f_{2}}$

$- \frac{f_{2}}{f_{1}}$

$- \frac{f_{2}}{f_{1} - f_{2}}$

$- \frac{f_{1}}{f_{1} + f_{2}}$

answer is B.

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Detailed Solution

u₁ = -x
$\begin{array}{l} \frac{1}{v_{1}} - \frac{1}{- x} = \frac{1}{f_{1}} \\ v_{1} = \frac{{xf}_{1}}{x - f_{1}} \\ m_{1} = \frac{v_{1}}{u_{1}} = \frac{v_{1}}{- x} = - (\frac{f_{1}}{x - f_{1}}) \end{array}$

u₂ = - (d - v₁)
$\begin{array}{l} \frac{1}{v_{2}} - \frac{1}{- (d - v_{1})} = \frac{1}{f_{2}} \\ v_{2} = \frac{(d - v_{1}) f_{2}}{d - v_{1} - f_{2}} \\ m_{2} = \frac{v_{2}}{- (d - v_{1})} = - (\frac{f_{2}}{d - v_{1} - f_{2}}) \end{array}$
$m_{1} m_{2} = (\frac{f_{1}}{x - f_{1}}) (\frac{f_{2}}{d - \frac{{xf}_{1}}{(x - f_{1})} - f_{2}}) = \frac{f_{1} f_{2}}{x (d - f_{1} - f_{2}) - {df}_{1} + f_{1} f_{2}}$
Since m is independent of x
$\begin{array}{l} \Rightarrow (d - f_{1} - f_{2}) = 0 \Rightarrow d = f_{1} + f_{2} \\ \Rightarrow m = - \frac{f_{2}}{f_{1}} \end{array}$