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Question

A convex lens of refractive index 32 has a power of 2.5 D in air. If it is placed in a liquid of refractive index 2, then the new power of the lens is

A
1.5 D
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B
1.2 D
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C
1.25 D
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D
1.25 D
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Solution

The correct option is C 1.25 D
Let the radius of curvature of the convex lens is R1 and R2

Focal length of a convex lens having power 2.5 D,=12.5 m

Focal length of a lens in air is given by,
1f=(μ1)(1R11R2)
2.5=1f=(321)(1R11R2).....(i)

Focal length of the lens in a liquid of refractive index 2 is
1f=⎜ ⎜ ⎜3221⎟ ⎟ ⎟(1R11R2)....(ii)
Dividing the equation (i) by equation (ii),
2.5f=0.50.25
1f=2.5×0.250.5=1.25 D

Hence, option (a) is correct.

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