Optics 7 Question 32

33. Consider a concave mirror and a convex lens (refractive index =1.5 ) of focal length 10cm each, separated by a distance of 50cm in air (refractive index =1) as shown in the figure. An object is placed at a distance of 15cm from the mirror. Its erect image formed by this combination has magnification M1.

(2015 Adv.) When the set-up is kept in a medium of refractive index 76, the magnification becomes M2. The magnitude M2M1 is

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Solution:

  1. Case I

Reflection from mirror

1f=1v+1u110=1v+115v=30

For lens

(in air)

|M1|=|v1u1||v2u2|=30152020=2×1=2

Case II For mirror, there is no change.

For lens, 1fair =3/2111R11R2

1fmedium =3/27/611R11R2

with

fair =10cm

We get

1fmedium =470cm1

1v120=4701v+120=27210=4701v=470120v=140|M2|=|v1u1||v2u2|=301514020=(2)14020=14M2M1=142=7



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