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A thin lens of index 1.52 has a focal length of 10 cm in air and 50 cm in a liquid. Find the index of the liquid.

User Kimbley
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1 Answer

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Final answer:

The index of the liquid is approximately 19.23.

Step-by-step explanation:

To find the index of the liquid, we can use the lens formula:

1/f = (n - 1)(1/R₁ - 1/R₂)

Where f is the focal length in meters, n is the refractive index of the lens, and R₁ and R₂ are the radii of curvature of the lens surfaces.

From the given information, we know that the focal length in air is 10 cm, so f = 0.10 m. We also know that the refractive index of the lens is 1.52. Plugging these values into the lens formula:

1/0.10 = (1.52 - 1)(1/∞ - 1/50)

Simplifying the equation gives:

10 = 0.52(0 - 1/50)

10 = -0.52/50

Solving for the refractive index of the liquid, we get:

1 - 10/0.52 = n - 1

n = 1 + 10/0.52

n ≈ 19.23

Therefore, the index of the liquid is approximately 19.23.

User Tsimtsum
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