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Chapter 34: Problem 47

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 is the (a) smaller and (b) larger radius?

Short Answer

Expert verified
The smaller radius \( R_1 \) is 240 mm, and the larger radius \( R_2 \) is 480 mm.

Step by step solution

01

Identify Known Values

We know the index of refraction of the glass is \( n = 1.5 \) and the focal length of the lens \( f = 60 \text{ mm} \). The lens maker's formula \( \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \) will be used, where \( R_2 = 2R_1 \) due to the problem condition.
02

Plug Known Values into Lens Maker's Equation

Starting with the lens maker's formula \( \frac{1}{60} = (1.5-1)\left(\frac{1}{R_1} - \frac{1}{2R_1}\right) \). Simplify the term \( \frac{1}{R_1} - \frac{1}{2R_1} = \frac{2-1}{2R_1} = \frac{1}{2R_1} \).
03

Solve for \( R_1 \)

With the equation \( \frac{1}{60} = 0.5 \times \frac{1}{2R_1} \), rewrite as \( \frac{1}{60} = \frac{1}{4R_1} \). To solve for \( R_1 \), rearrange to get \( R_1 = 4 \times 60 = 240 \text{ mm} \).
04

Calculate \( R_2 \)

Given \( R_2 = 2R_1 \), substitute the value of \( R_1 \) to find the larger radius. Thus, \( R_2 = 2 \times 240 = 480 \text{ mm} \).

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Key Concepts

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

Radius of Curvature
In lens optics, the radius of curvature refers to the radius of the sphere from which a lens surface is shaped. A lens typically has two surfaces - each can be convex, concave, or a combination of these.
For a double-convex lens, both surfaces are outward curved. The radius of curvature affects how the lens bends light, a fundamental characteristic determining its focal length.
  • The smaller the radius, the "sharper" the curve of the lens.
  • A larger radius implies a "flatter" curve.
In the original exercise, one surface has twice the radius of curvature of the other. This indicates that one surface is less curved than the other, meaning it bends light less sharply. Understanding these properties is important as they affect how well the lens can focus light, ultimately influencing the focal length.
Lens Maker's Formula
Lens maker's formula is crucial in determining how lenses focus light. This formula relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces.
The formula is expressed as:\[\frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]where:
  • \(f\) is the focal length of the lens,
  • \(n\) is the refractive index of the glass used to make the lens,
  • \(R_1\) and \(R_2\) are the radii of curvature of the two lens surfaces.
This equation helps in designing lenses with desired focal lengths by adjusting the shape and material of the lens. In the provided exercise, using the known values and relation \(R_2 = 2R_1\) simplifies calculations, ultimately enabling the calculation of both radii.
Focal Length Calculation
The focal length of a lens is the distance from the center of the lens to the focal point, where parallel rays of light converge. It is a critical measure of how strongly a lens converges or diverges light.
For the exercise at hand, the desired focal length is specified as \(60 \text{ mm}\), guiding the choice of the lens's radii of curvature. Using known values and the lens maker's formula, the focal length informs adjustments to \(R_1\) and \(R_2\).
The step-by-step process in the workout leads to:
  • Determining \( R_1=240 \text{ mm} \): As the rearranged equation shows \( R_1 \) to be directly proportional to the focal length, solve for it by isolating \( R_1 \).
  • Calculating \( R_2 \): By simply doubling the value of \( R_1 \) since \( R_2 = 2R_1 \), we find \( R_2 = 480 \text{ mm} \).
Understanding these calculations highlights the integral role of carefully selecting dimensions to attain desired optical outcomes.

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Most popular questions from this chapter

Two thin lenses of focal lengths \(f_{1}\) and \(f_{2}\) are in contact and share the same central axis. Show that, in image formation, they are equivalent to a single thin lens for which the focal length is \(f=f_{1} f_{2} /\left(f_{1}+f_{2}\right)\)

You look through a camera toward an image of a hummingbird in a plane mirror. The camera is \(4.30 \mathrm{~m}\) in front of the mirror. The bird is at camera level, \(5.00 \mathrm{~m}\) to your right and \(3.30 \mathrm{~m}\) from the mirror. What is the distance between the camera and the apparent position of the bird's image in the mirror?

Light travels from point \(A\) to point \(B\) via reflection at point \(O\) on the surface of a mirror. Without using calculus, show that length \(A O B\) is a minimum when the angle of incidence \(\theta\) is equal to the angle of reflection \(\phi .\) (Hint: Consider the image of \(A\) in the mirror.

An object is \(30.0 \mathrm{~cm}\) from a spherical mirror, along the mirror's central axis. The mirror produces an inverted image with a lateral magnification of absolute value \(0.500 .\) What is the focal length of the mirror?

(a) A luminous point is moving at speed \(v_{O}\) toward a spherical mirror with radius of curvature \(r\), along the central axis of the mirror. Show that the image of this point is moving at speed \(v_{I}=-\left(\frac{r}{2 p-r}\right)^{2} v_{O}\) where \(p\) is the distance of the luminous point from the mirror at any given time. Now assume the mirror is concave, with \(r=15 \mathrm{~cm}\), and let \(v_{O}=5.0 \mathrm{~cm} / \mathrm{s} .\) Find \(v_{I}\) when \((\mathrm{b}) p=30 \mathrm{~cm}\) (far outside the focal point \(),(\mathrm{c}) p=8.0 \mathrm{~cm}\) (just outside the focal point), and (d) \(p=10 \mathrm{~mm}\) (very near the mirror).
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