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

25 c m

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15 c m

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30 c m

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50 c m

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Solution

The correct option is B $15 c m$
$P = \frac{1}{f_{1}} = \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}$
$0 = \frac{1}{20} - \frac{1}{5} - \frac{d}{20 (- 5)}$
$\frac{d}{100} = \frac{1}{5} - \frac{1}{20} = \frac{4 - 1}{20} = \frac{3}{20}$
or $d = 15 c m$ .

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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)

Q. 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 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 G of focal length 5 cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance d in cm will be

The correct option is B 15cmP=1f1=1f2−df1f20=120−15−d20(−5)d100=15−120=4−120=320or d=15 cm.

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

The correct option is B $15 c m$
$P = \frac{1}{f_{1}} = \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}$
$0 = \frac{1}{20} - \frac{1}{5} - \frac{d}{20 (- 5)}$
$\frac{d}{100} = \frac{1}{5} - \frac{1}{20} = \frac{4 - 1}{20} = \frac{3}{20}$
or $d = 15 c m$ .