Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d where d - f1- d - f2- and their centre are displaced by a vertical separation - as shown in figure- Taking the origin of coordinates- O- at the centre 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

The correct option is C x= f _ 1 f _ 2 +d( f _ 1 d ) f _ 1 + f _ 2 d , y=Δ( f _ 1 d ) f _ 1 + f _ 2 d For lens 2 u amp;= (d f_1) ...

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Byju's Answer

Standard XII

Physics

Combinations of Lenses & Mirrors

Two thin conv...

Question

Two thin convex lenses of focal lengths $f_{1}$ and $f_{2}$ are separated by a horizontal distance $d$ (where $d > f_{1}, d > f_{2}$ ), and their centre are displaced by a vertical separation $Δ$ as shown in figure. Taking the origin of coordinates, $O$ , at the centre 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

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

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

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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}

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

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Solution

The correct option is C $x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = \frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d}$
$For lens - 2$

$\begin{matrix} u & = - (d - f_{1}) h_{0} & = - Δ f & = f_{2} \end{matrix}$

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

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

$\frac{1}{v} = \frac{1}{f_{2}} - \frac{1}{d - f_{1}}$

$v = \frac{f_{2} (d - f_{1})}{d - f_{1} - f_{2}} =$ distance from lens $_{2} .$

Distance form O (Lens-1) is

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

$= \frac{d^{2} - d f_{1} - d f_{2}^{'} + f_{2} A - f_{2} f_{1}}{d - f_{1} - f_{2}}$

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

magnification $= \frac{h_{1}}{h_{0}} = \frac{v}{u}$

$\begin{matrix} = \frac{h_{1}}{Δ} & = \frac{f_{2} (d - f_{1})}{\frac{d - f_{1} - f_{2}}{- (d - f_{1})}} y_{1} & = \frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d} \end{matrix}$

So option $c$ is correct.

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Similar questions

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