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Chapter 31: Problem 26

You're writing specifications for a new line of magnifying glasses that have double-convex lenses with equal 32 -cm curvature radii, made from glass with \(n=1.52 .\) What do you list for the focal length?

Short Answer

Expert verified
The focal length of the magnifying glasses is approximately 48 cm.

Step by step solution

01

Understand The Lensmaker's Equation

The Lensmaker's Equation is a formula that relates the focal length of a lens with its radii of curvature and refractive index. The equation 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 refractive index of the lens material, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.
02

Adapt The Lensmaker's Equation To The Specifics Of The Problem

Since the lens is a double-convex lens with equal curvature radii, both curvature radii are the same (denote it as \( R \)). Moreover, for a double-convex lens, if we consider the incident light to come from the left the signs for \( R_1 \) and \( R_2 \) are positive and negative respectively. Hence, in this scenario, \( R_1 = R \) and \( R_2 = -R \). Substituting these values into the lensmaker's equation, we obtain \[ \frac{1}{f} = (n-1) \left( \frac{1}{R} + \frac{1}{R} \right) = 2(n-1)/R \].
03

Calculate the Focal Length

With \( R = 32 \) cm and \( n = 1.52 \), substituting these values into the adapted equation gives: \[ \frac{1}{f} = 2 * (1.52 - 1) / 32 \] Solve for \( f \) to find the focal length.

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Most popular questions from this chapter

By holding a magnifying glass \(25 \mathrm{cm}\) from your desk lamp, you can focus an image of the lamp's bulb on a wall \(1.6 \mathrm{m}\) from the lamp. What's the focal length of your magnifying glass?

The cornea of the human eye has refractive index \(1.38,\) while the eye's lens has a graduated index in the range 1.38 to \(1.40 ;\) use 1.39 for this problem. For the aqueous humor between cornea and lens, \(n=1.34 .\) Find the angle through which light is deflected at the first surface of (a) the cornea and (b) the lens, if it's incident at \(20^{\circ}\) to the normal at each surface. Your result shows that the cornea is the dominant refractive element in the eye.

An object is five focal lengths from a concave mirror. (a) How do the object and image heights compare? (b) Is the image upright or inverted?

For what refractive index would the focal length of a plano-convex lens be equal to the curvature radius of its one curved surface?

A particular eye has a focal length of \(2.0 \mathrm{cm}\) instead of the \(2.2 \mathrm{cm}\) that would put a sharply focused image on the retina. (a) Is this eye nearsighted or farsighted? (b) What corrective lens is needed?
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