JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Geometrical Optics

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The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm . The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be

x : 1

. The value of x is ____ .

Answer: 4

Solution

By the lens maker's formula the power in air is

P = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

, while in a liquid the factor

(n - 1)

is replaced by

(\frac{n}{n _{l i q}} - 1)

with the surface term unchanged. The ratio therefore depends only on these factors:

\frac{P _{ai r}}{P _{l i q}} = \frac{n - 1}{\frac{n}{n _{l i q}} - 1} = \frac{1.8 - 1}{\frac{1.8}{1.5} - 1} = \frac{0.8}{0.2} = 4

. Hence

x = 4

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer

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