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Chapter 35: Problem 39

A double-convex lens is to be made of glass with an index of refraction of \(1.5 .\) One surface is to have twice the radius of curvature of the other and the focal length is to be \(60 \mathrm{~mm}\). What are the radii?

Short Answer

Expert verified
The radii are 15 mm and 30 mm.

Step by step solution

01

- Understand the Lensmaker's Equation

The Lensmaker's Equation for a double-convex lens is: \[\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]Where: - \(f\) is the focal length. - \(n\) is the index of refraction.- \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
02

- Assign Variables

Let the radius of curvature for the first surface be \(R_1\). Therefore, the radius of curvature for the second surface \(R_2\) will be \(2R_1\). The focal length \(f\) is given as 60 mm, and the index of refraction \(n\) is 1.5.
03

- Substitute into Lensmaker鈥檚 Equation

Substitute the given values into the Lensmaker鈥檚 Equation:\[\frac{1}{60} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{2R_1} \right)\]
04

- Simplify the Equation

Simplify the equation as follows:\[\frac{1}{60} = 0.5 \left( \frac{1}{R_1} - \frac{1}{2R_1} \right)\]Combine the fractions inside the parentheses:\[\frac{1}{60} = 0.5 \cdot \frac{2 - 1}{2R_1} = 0.5 \cdot \frac{1}{2R_1} = \frac{1}{4R_1}\]
05

- Solve for \(R_1\)

Solve the equation for \(R_1\):\[\frac{1}{60} = \frac{1}{4R_1} \implies 4R_1 = 60 \implies R_1 = 15 \text{ mm}\]
06

- Find \(R_2\)

Since \(R_2\) is twice \(R_1\):\[R_2 = 2R_1 = 2 \times 15 = 30 \text{ mm}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

focal length
The focal length is a crucial property of lenses. It represents the distance between the lens and the point where light rays converge to form a sharp image. In the context of the Lensmaker's Equation, it's symbolized as 'f'. For a double-convex lens, this equation tells us how the radii of curvature and the material's index of refraction impact the focal length. With a focal length of 60mm given in our problem, it becomes the key parameter to solve for the radii of curvature of the lens surfaces.
index of refraction
The index of refraction (n) quantifies how much a material slows down light passing through it. It's the ratio of the speed of light in a vacuum to its speed in the medium. For the given problem, the glass has an index of refraction of 1.5. This means that light travels 1.5 times slower in glass than in a vacuum. In the Lensmaker's Equation, the index of refraction is a critical factor affecting how the lens bends light. Remember, a higher index of refraction means more bending of light.
radii of curvature
The radii of curvature, denoted as R1 and R2 in the Lensmaker's Equation, are the radii of the spherical surfaces of the lens. For a double-convex lens, one radius of curvature determines the other since one surface is twice the radius of the other. The given exercise sets R2 to be twice R1. By solving the Lensmaker's Equation with the given values, we find R1 to be 15 mm and R2 to be 30 mm. These values directly influence how the lens converges light to achieve the given focal length of 60 mm.

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Most popular questions from this chapter

The index of refraction of benzene is \(1.8\). What is the critical angle for a light ray traveling in benzene toward a plane layer of air above the benzene?

When a T. rex pursues a jeep in the movie Jurassic Park, we see a reflected image of the (very large) T. rex via a side-view mirror, on which is printed the (then darkly humorous) warning: "Objects in mirror are closer than they appear." Is the mirror flat, convex, or concave? Why do you think so?

A simple magnifying lens of focal length \(\bar{f}\) is placed near the eye of someone whose near point \(P_{n}\) is \(25 \mathrm{~cm}\) from the eye. An object is positioned so that its image in the magnifying lens appears at \(P_{n} .\) (a) What is the lens's angular magnification? (b) What is the angular magnification if the object is moved so that its image appears at infinity? (c) Evaluate the angular magnifications of (a) and (b) for \(f=10 \mathrm{~cm}\). (Viewing an image at \(P_{n}\) requires effort by muscles in the eye, whereas for many people viewing an image at infinity requires no effort.)

In about A.D. 150 , Claudius Ptolemy gave the following measured values for the angle of incidence \(\theta_{1}\) and the angle of refraction \(\theta_{2}\) for a light beam passing from air to water: $$\begin{array}{cc|cc} \hline \boldsymbol{\theta}_{\mathbf{1}} & \boldsymbol{\theta}_{\mathbf{2}} & \boldsymbol{\theta}_{\mathbf{1}} & \boldsymbol{\theta}_{\mathbf{2}} \\ \hline 10^{\circ} & 8^{\circ} 00^{\prime} & 50^{\circ} & 35^{\circ} 00^{\prime} \\ 20^{\circ} & 15^{\circ} 30^{\prime} & 60^{\circ} & 45^{\circ} 30^{\prime} \\ 30^{\circ} & 22^{\circ} 30^{\prime} & 70^{\circ} & 45^{\circ} 30^{\prime} \\ 40^{\circ} & 29^{\circ} 00^{\prime} & 80^{\circ} & 50^{\circ} 00^{\prime} \\ \hline \end{array}$$ (a) Are these data consistent with the law of refraction? (b) If so, what index of refraction results? These data are interesting as perhaps the oldest recorded physical measurements.

Two thin lenses of focal lengths \(f_{1}\) and \(f_{2}\) are in contact. Show that they are equivalent to a single thin lens with $$f=\frac{f_{1} f_{2}}{f_{1}+f_{2}}$$ as its focal length.
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