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Question

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 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 of M2M1 is

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Solution

From the mirror formula,
1f=1v+1u,u=15 cm,f=10 cm
110=1v+115
So, v=30cm (right).
Magnification due to mirror is M1m=vu=2
Now u for the lens =20 cm=2flens.
From the lens formula , vlens=20cm (right)
magnification by the lens |M1L|=1(in air).
Net Magnification M1=2(1)=2

When kept in liquid (μ=76),
Change of medium does not affect f for mirror and hence magnification as well.
M2m=2
The focal length of the lens is changed from f=10 cm to f
1f1f=(μg1)(1R11R2)(μgμ1)(1R11R2)=(321)(32671)f=352cm.

So, for lens, u=20 cm and 1v130=235v=140 cm.
M2L=vu=14020=7.
Net magnification M2=2(7)=14
Now, for the system, M2M1=7.

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