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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 cm

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- 30 cm

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- 15 cm

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- 60 cm

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Solution

The correct option is D $- 60 cm$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, f = + 15 cm, u = - 40 cm$
$\frac{1}{v} - \frac{1}{- 40} = \frac{1}{15}, v = + 24 cm$
This image acts as object for lens $L_{2}$
So, $u = + (24 - 4) = + 20 cm$
$v = 30 cm$
$f = ?$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{+ 30} - \frac{1}{+ 20}$
$f = - 60 cm$

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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 D −60 cm1v−1u=1f,f=+15 cm,u=−40 cm 1v−1−40=115,v=+24 cm This image acts as object for lens L2 So, u=+(24−4)=+20 cm v=30 cm f=? 1f=1v−1u⇒1f=1+30−1+20 f=−60 cm

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 $L_{2}$ .