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>>Combination of Mirror and Lenses
>> $A biconvex lens of focal length 15 cm is$

Question

A biconvex lens of focal length $15 c m$ is in front of a plane mirror.The distance between lens and the mirror is $10 c m .$ A small object is kept at a distance of $30 c m$ from the lens. The final image is:

A

virtual and at a distance of $16 c m$ from the mirror

B

real and at a distance of $16 c m$ from the mirror

C

virtual and at a distance of $20 c m$ from the mirror

D

real and at a distance of $20 c m$ from the mirror

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Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is B)

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{1 5} = \frac{1}{v} - \frac{1}{- 3 0} v = + 30 \frac{1}{1 5} = \frac{1}{v} - \frac{1}{1 0} \Rightarrow \frac{1}{1 5} + \frac{1}{1 0} = \frac{1}{v} v = \frac{1 5 0}{2 5} = + 6 c m$

The final image $I_{3}$ is $6 c m$ (real) from
the lens.
$10 + 6 = 16 c m$
from the mirror ( real)
So option $B$ is correct.

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