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A contact lens is made of plastic with an index of refraction of 1.60. The lens has an outer radius of curvature of +2.10 cm and an inner radius of curvature of +2.44 cm. What is the focal length of the lens?

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Answer: approximately 25.25 cm.

Explanation:

To find the focal length of the lens, we can use the lens maker's formula, which relates the focal length (f) of a lens to the radii of curvature (R1 and R2) of its surfaces and the index of refraction (n) of the material.

The lens maker's formula is given by:

1/f = (n - 1) * (1/R1 - 1/R2)

Given:

Index of refraction (n) = 1.60

Outer radius of curvature (R1) = +2.10 cm

Inner radius of curvature (R2) = +2.44 cm

Substituting the given values into the formula, we get:

1/f = (1.60 - 1) * (1/2.10 - 1/2.44)

Simplifying the equation:

1/f = 0.60 * (0.476 - 0.410)

1/f = 0.60 * 0.066

1/f = 0.0396

Now, we can solve for f by taking the reciprocal of both sides:

f = 1 / 0.0396

f ≈ 25.25 cm

Therefore, the focal length of the contact lens is approximately 25.25 cm.

User Rakesh Chouhan
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