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A Convex Lens of Refractive Index 1.5 Has a Focal Length of 18 Cm in Air .Calculate the Change in Its Focal Length When It is Immersed in Water of Refractive Index 4 3 . - Physics

Question

A convex lens of refractive index 1.5 has a focal length of 18 cm in air .Calculate the change in its focal length when it is immersed in water of refractive index `4/3`.

Solution

Reflective index of convex lens in air, μ = 1.5

Focal length of convex lens, fa = 18 cm

Reflective index of water, μ′ = `4/3`

When the lens is in air:

`I/f_a  = (mu - 1) (1/R_1 - 1/R_2)`

Where R1 and Rare the radius of curvature of the convex lens

`1/18 = (1.5 - 1) (1/R_1  - 1/R_2)`

`(1/R_1  - 1/R_2)  = 1/9`

When the lens is in air:

`1/f_w  = (mu' - 1) (1/R_1  -1/ R_2)`

Where, fw is the focal length of the lens, when immersed in water

`1/f_w  = ((1.5)/(4/3)  - 1 ) (1/R_1  - 1/R_2)`

`1/f_w  = 1/8 xx 1/9`

`f_w =72 cm`

Change in focal length = f− f= 72 cm − 18 cm = 54 cm

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A Convex Lens of Refractive Index 1.5 Has a Focal Length of 18 Cm in Air .Calculate the Change in Its Focal Length When It is Immersed in Water of Refractive Index 4 3 . Concept: Lenses.
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