A convex lens $A$ of focal length $20 c m$ and a concave lens $B$ of focal length $5 c m$ are kept along the same axis with a distance $d$ between them. If a parallel beam of light falling on $A$ leaves $B$ as a parallel beam, then $d$ is equal to ________ cm$.

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Solution

$P_{1} = \frac{100}{20} = + 5 P_{2} = \frac{- 100}{5} = - 20 0 = P_{1} + P_{2} + d P_{1} P_{2} 0 = 5 - 20 - 100 d \Rightarrow d = \frac{- 15}{100}$
$d = 15$ cm

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A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distace d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then d is equal to
(IIT-JEE 1985)

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A

of focal length

20 c m

and a concave lens

B

of focal length

5 c m

are kept along the same axis with a distance

^{'} d^{'}

between them as shown in a figure. If a parallel beam falling on

A

emerges parallel from

B

, then

^{'} d^{'}

is equal to

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A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then d is equal to ________ cm$.

P1=10020=+5P2=−1005=−200=P1+P2+dP1P20=5−20−100d⇒d=−15100d=15cm

A convex lens $A$ of focal length $20 c m$ and a concave lens $B$ of focal length $5 c m$ are kept along the same axis with a distance $d$ between them. If a parallel beam of light falling on $A$ leaves $B$ as a parallel beam, then $d$ is equal to ________ cm$.

$P_{1} = \frac{100}{20} = + 5 P_{2} = \frac{- 100}{5} = - 20 0 = P_{1} + P_{2} + d P_{1} P_{2} 0 = 5 - 20 - 100 d \Rightarrow d = \frac{- 15}{100}$
$d = 15$ cm