JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Geometrical Optics

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A convex lens of refractive index 1.5 and focal length 18 cm in air is immersed in water. The change in focal length of the lens will be ______ cm.
(Given refractive index of water

= \frac{4}{3}

)

Answer: 54

Solution

The lens maker's formula in air gives

\frac{1}{f _{ai r}} = (μ_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

. When the lens is immersed in water the factor

(μ_{g} - 1)

is replaced by

(\frac{μ _{g}}{μ _{w}} - 1)

, the surface curvature being unchanged, so

\frac{f _{w a t er}}{f _{ai r}} = \frac{μ _{g} - 1}{\frac{μ _{g}}{μ _{w}} - 1} = \frac{0.5}{\frac{1.5}{4/3} - 1} = \frac{0.5}{0.125} = 4

. Hence

f_{w a t er} = 4 \times 18 = 72

cm, and the focal length increases by

72 - 18 = 54

cm.

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Integer answer

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