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Combinations of Lenses & Mirrors

In the adjoin...

Question

In the adjoining ray - diagram are given the positions of an object O, image I and two lenses $L_{1}$ and $L_{2}$ . The focal length of $L_{1}$ is also given. Find the focal length of the lens $L_{2}$

30 c m

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- 30 c m

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- 15 c m

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None of these

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Solution

The correct option is C $- 15 c m$
For lens $L_{1}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f},$
$f = + 15 c m u = - 40 c m$
put these values
$\frac{1}{v} - \frac{1}{- 40} = \frac{1}{15}, v = + 24 cm$
This image acts as object for lens $L_{2}$
So, $u = + (24 - 14) = + 10 c m$ ,
$v = 30 c m$
$f = ?$
put the values
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{+ 30} - \frac{1}{+ 10}$
$f = - 15 c m$

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In the adjoining ray - diagram are given the positions of an object O, image I and two lenses L1 and L2. The focal length of L1 is also given. Find the focal length of the lens L2

The correct option is C −15 cmFor lens L1 1v−1u=1f, f=+15 cmu=−40 cm put these values 1v−1−40=115,v=+24 cm This image acts as object for lens L2 So, u=+(24−14)=+10 cm, v=30 cm f=? put the values 1f=1v−1u⇒1f=1+30−1+10 f=−15 cm

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