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Question

The radii of curvature of a lens are + 20 cm and + 30 cm. The material of the lens has a refracting index 1.6. Find the focal length of the lens (a) if it is placed in air, and (b) if it is placed in water (μ = 1.33).

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Solution

Given,
Radii of curvature of a lens, (R1) = +20 cm and R2 = +30 cm
Refractive index of the material of the lens, (μ) = 1.6
Refractive index of water, (μwater) = 1.33

(a)
When the lens is placed in air,
Using lens maker formula:

1f=(μ-1)1R1-1R2
1f=0.6120-130
f=0.61060×10
f = 100 cm
Thus, the focal length of the lens is 100 cm when it is placed in air.

(b)When the lens is placed in water

1f=μlensμwater-11R1-1R2
=1.601.33-1160
=28133×601300
f = 300 cm.
Thus, the focal length of the lens is 300 cm when it is placed in water.


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