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Chapter 18: Problem 373

Find the ratio of the focal lengths of a glass lens in water and in air. The refractive indices of the glass and water are \(1.5\) and \(1.33\) respectively.

Short Answer

Expert verified
The ratio of the focal lengths of a glass lens in water and in air is approximately \(1.47\).

Step by step solution

01

Recall Lens Maker's Formula

The lens maker's formula relates the focal length of a lens (f) to the refractive index of the lens material (n) and the radii of curvature of the lens surfaces (R1 and R2): \( \frac{1}{f} = (n - 1)(\frac{1}{R_1} - \frac{1}{R_2}) \)
02

Find the Focal Length in Air

When the lens is in air, the refractive index of the surroundings is 1. Let n1 be the refractive index of glass (1.5). Let f1 be the focal length of the glass lens in air, so we have: \( \frac{1}{f_1} = (n_1 - 1)(\frac{1}{R_1} - \frac{1}{R_2}) \)
03

Find the Focal Length in Water

When the lens is in water, the refractive index of the surroundings is 1.33. We need to find the relative refractive index of glass with respect to water. Let n2 be the relative refractive index of the glass lens in water: \( n_2 = \frac{n_1}{n_{water}} = \frac{1.5}{1.33} \) Now, let f2 be the focal length of the glass lens in water, so we have: \( \frac{1}{f_2} = (n_2 - 1)(\frac{1}{R_1} - \frac{1}{R_2}) \)
04

Find the Ratio of Focal Lengths

Now, we need to find the ratio of the focal lengths in water and in air: \( \frac{f_1}{f_2} = \frac{(n_1 - 1)(\frac{1}{R_1} - \frac{1}{R_2})}{(n_2 - 1)(\frac{1}{R_1} - \frac{1}{R_2})} \) Since \(\frac{1}{R_1} - \frac{1}{R_2}\) is common in both numerator and denominator, it gets canceled out: \( \frac{f_1}{f_2} = \frac{n_1 - 1}{n_2 - 1} \) Now, substitute the values of n1 (=1.5) and n2: \( \frac{f_1}{f_2} = \frac{1.5 - 1}{\frac{1.5}{1.33} - 1} \)
05

Calculate the Ratio

Calculate the ratio of the focal lengths: \( \frac{f_1}{f_2} = \frac{0.5}{\frac{1.5}{1.33} - 1} \approx 1.47 \) So, the ratio of the focal lengths of the glass lens in water and in air is approximately 1.47.

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Most popular questions from this chapter

Find the ratio of the focal lengths of a glass lens in water and in air. The refractive indices of the glass and water are \(1.5\) and \(1.33\) respectively.

An axial point object is located \(25 \mathrm{~cm}\) to the left of a thin lens of focal length \(+10.00 \mathrm{~cm}\). A ray of light coming from the object subtends an angle of \(10^{\circ}\) with the axis. At what angle does this ray, after refraction, intersect the axis?

An object \(4 \mathrm{~cm}\) in diameter is placed \(167 \mathrm{~mm}\) from a converging lens of 5 diopters. Calculate the position of the image from the lens and its size. Is the image real or virtual?

If the two principal points of a lens surrounded by the same medium on both sides coincide with each other at a point midway between the two vertices, what is the form of the lens?

If two thin lenses are placed on the same axis with their second focal points in coincidence, show that the second focal point of the combination is midway between the common focal point and the second lens, and that the deviation produced by the second lens is twice that produced by the first lens (assuming that the angles are small).
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