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Question

Two identical plano-convex lenses L1(μ1=1.4) and L2(μ2=1.5) of radii of curvature R=20 cm are placed as shown in the figure.
Now, the second lens is shifted vertically downwards by a small distance 4.5 mm and the extended parts of L1 and L2 are blackened as shown in figure. Find the new position of the image of the parallel beam.
160959_28f0cf03d35e4c14aa0a7324a425e595.png
  1. 100/9cm behind the lens 2mm below the principal axis of L1
  2. 200/9cm behind the lens 2.5mm below the principal axis of L1
  3. 200/9cm in front of the lens 2.5mm below the principal axis of L1
  4. 100/9cm in front of the lens 2mm below the principal axis of L1

A
200/9cm behind the lens 2.5mm below the principal axis of L1
B
100/9cm behind the lens 2mm below the principal axis of L1
C
200/9cm in front of the lens 2.5mm below the principal axis of L1
D
100/9cm in front of the lens 2mm below the principal axis of L1
Solution
Verified by Toppr

Consider refraction of light from first surface of L1,

1.4v11=1.4120

Consider refraction of light from second surface of L1,

1v21.4v1=11.4

Hence, v2=50cm

Now this image is obtained after refraction from lens L1 and lies on the principal axis of L1.

This image gets refracted from L2 for which it is at a height of
4.5mm.

Consider refraction of light from first surface of L2,

1.5v31v2=1.51

Consider refraction of light from second surface of L2,

1v41.5v3=11.520

v4=2009cm

Height of final image =M×4.5mm=2mm

This is the height of object above principal axis of L2.

Hence the final image is 4.5mm2mm=2.5mm below principle axis of
L1.

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