Question

A lens of a pair of eyeglasses with index of refraction 1.723 has a power of \(4.29 \mathrm{D}\) in air. What is the power of this lens if it is put in water with \(n=1.333 ?\)

Short answer

Answer: The power of the lens when it is put in water is approximately 3.39D.

Step-by-step solution

Find the focal length of the lens in air in meters

To find the focal length (in meters), we'll use the formula: Focal length (air) = 1/Power (air) Given that the power of the lens in air is 4.29D, we can calculate the focal length as follows: Focal length (air) = \(\frac{1}{4.29}\) m

Find the focal length of the lens in water

To find the focal length of the lens in water, we will begin by using the lens-maker's equation: \(\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\) Where f is the focal length, n is the index of refraction of the lens, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces. Since we don't have information about the radii of curvature, we can use the relationship between the focal length in different media: \(\frac{f_{water}}{f_{air}} = \frac{n_{lens} - n_{water}}{n_{lens} - n_{air}}\) Plugging in the values we have: \(\frac{f_{water}}{\frac{1}{4.29}} = \frac{1.723 - 1.333}{1.723 - 1}\) Now, we can solve for \(f_{water}\): \(f_{water} = \frac{1}{4.29} \times \frac{1.723 - 1.333}{1.723 - 1}\)

Calculate the power of the lens in water

Now that we have the focal length of the lens in water, we can calculate its power using the formula: Power (water) = \(\frac{1}{f_{water}}\) Plugging in the value of \(f_{water}\), we get: Power (water) = \(\frac{1}{\left( \frac{1}{4.29} \times \frac{1.723 - 1.333}{1.723 - 1} \right)}\)

Simplify the expression

Now we can simplify the expression for Power (water): Power (water) = \(\frac{4.29}{\frac{1.723 - 1.333}{1.723 - 1}}\) Power (water) = \(4.29 \times \frac{1.723 - 1}{1.723 - 1.333}\) Power (water) ≈ 3.39 D Thus, the power of the lens when it is put in water is approximately 3.39D.