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Question

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is :
  1. real and at a distance of 16 cm from the mirror
  2. virtual and at a distance of 16 cm from the mirror
  3. virtual and at a distance of 20 cm from the mirror
  4. real and at a distance of 20 cm from the mirror

A
virtual and at a distance of 16 cm from the mirror
B
virtual and at a distance of 20 cm from the mirror
C
real and at a distance of 16 cm from the mirror
D
real and at a distance of 20 cm from the mirror
Solution
Verified by Toppr

We get the image distance due to the lens using lens formula,

1v1u=1f

or

1v=115130

or

v=30cm in front of the convex mirror, which means 20cm behind the mirror. Assuming mirror to reflect from both the sides and this image will act as virtual object for mirror and second image, I2 will be formed at 20 cm in front of mirror. That is 10 cm in front of lens.

Understanding the convention used while taking +ve and -ve sign for image and object distance, the direction along which ray of light travels is taken as +ve.

So, here, the ray of light is coming from behind the mirror towards lens, for the second image, I2 with respect to lens,

u=+10cm;f=15cm

So, applying lens formula,

1v1u=1f

or

1v110=115

or

v=6cm from lens (in front of lens)

So this image is at a distance of 16 (10 + 6) cm in front of mirror.

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