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Question: Answered & Verified by Expert
Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1), as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image, formed by this combination, has magnification M1 . When the set-up is kept in a medium of refractive index 76 , the magnification becomes M2. The magnitude M2M1 is

PhysicsRay OpticsNEET
Options:
  • A 7
  • B 5
  • C 3
  • D 6
Solution:
2642 Upvotes Verified Answer
The correct answer is: 7
For reflection from a concave mirror,

1v+1u=1f1v-115= -110

1v=115-110=-130

v=-30

Magnification m1=-vu=-2

Now for refraction from lens,

1v-1u=1f1v=110-120=120

Magnification m2=vu=-1

M1=m1m2=2 

Now when the set-up is immersed in liquid, no effect for the image formed by mirror.

We have μL-11R1-1R2=110

1R1-1R2=15

When lens is immersed in liquid,

1flens=μLμS-1 1R1-1R2=27×15=235

1v-1u=1fLiquid

1v=235-120=8-7140=1140

Magnification =-14020=-7

M2=2×7=14

M2M1=7

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