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Chapter 17: Problem 99

Think \& Calculate A friend tells you that when he takes off his eyeglasses and holds them \(21 \mathrm{~cm}\) above a printed page, the image of the print is upright but reduced to \(0.67\) of its actual size. (a) Are the lenses in the glasses concave or convex? Explain. (b) What is the focal length of your friend's glasses?

Short Answer

Expert verified
(a) The lenses are concave. (b) The lenses have a focal length of approximately -14.08 cm.

Step by step solution

01

Determine Lens Type

Since the image is upright and reduced, the lenses must be concave. Concave lenses produce virtual, upright images that are smaller than the object.
02

Use the Magnification Formula

The magnification, given by \( m = \frac{h_{\text{image}}}{h_{\text{object}}} \), is 0.67. This implies that the image height is 67% of the object height.
03

Magnification and Lens Formula Relations

For lenses, magnification \( m = \frac{v}{u} \) where \( v \) is the image distance and \( u \) is the object distance. Here, \( m = 0.67 \) and \( u = -21 \) cm (negative because the object is real and on the same side as the light's approach). Therefore, \( v = -0.67 \times 21 \).
04

Calculate Image Distance

Calculate \( v \): \( v = -0.67 \times 21 \approx -14.07 \) cm. The negative sign confirms the virtual image, meaning it's on the same side as the object.
05

Lens Formula Application

Apply the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). We have \( v = -14.07 \) cm and \( u = -21 \) cm. Substitute these into the formula to find \( f \).
06

Calculate Focal Length

Substitute the known values: \( \frac{1}{f} = \frac{1}{-14.07} - \frac{1}{-21} \). Calculating gives \( \frac{1}{f} \approx -0.07099 \), so \( f \approx -14.08 \) cm.

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Key Concepts

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

Lens Magnification
Magnification in lenses is a measure of how much larger or smaller the image is compared to the original object. Concave lenses, specifically, produce an image that is smaller than the object. This is understood using the magnification formula:
  • Magnification, \( m = \frac{h_{\text{image}}}{h_{\text{object}}} \)
  • In the exercise, the magnification was given as 0.67, meaning the image size is 67% that of the object size.
  • This makes the image appear smaller when you look through the lens at the object.

A critical point to note is that with concave lenses, the images formed are typically virtual and upright, in addition to being diminished. This characteristic helps identify that the friend's eyeglasses are indeed concave because an amplified or equal-sized image feature would suggest a different lens type.
Focal Length Calculation
Calculating the focal length of a lens is crucial in understanding its optical power and how it focuses light. The exercise involves the lens formula, which is:
  • \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
  • Here, \( u \) is the object distance (-21 cm, negative for a real object), and \( v \) is the image distance.
  • We found \( v \) from the magnification formula as approximately -14.07 cm.

By substituting these values into the lens formula, \( \frac{1}{-14.07} - \frac{1}{-21} \), we derive \( f \), the focal length of the lens.

Finally, solving for \( f \) provides approximately -14.08 cm. The negative result indicates that the lens is concave, since concave lenses have a negative focal length.
Image Formation
Understanding how an image is formed through lenses is pivotal to optics. With concave lenses, the image formed has distinct characteristics:
  • It is virtual, meaning it cannot be projected onto a screen because the light rays do not actually converge in reality.
  • The image is diminished, smaller than the actual object, which aligns with the magnification of 0.67 in this situation.
  • The image is upright, meaning it has the same orientation as the object.

Image formation through concave lenses can be visualized as diverging light rays. When parallel rays of light pass through the lens, they spread out, making it appear as though they are coming from a point behind the lens. This is why such lenses are used in eyeglasses for people with myopia, to correct the focus by causing the light to diverge before reaching their eyes, aligning the focal plane with the retina.

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Most popular questions from this chapter

Think \& Calculate An object is located to the left of a convex lens whose focal length is \(36 \mathrm{~cm}\). The magnification produced by the lens is \(m=3.0\). (a) To increase the magnification to \(4.0\), should the object be moved closer to the lens or farther away? Explain. (b) Find the object distance that gives a magnification of \(4.0\).

ptolemy鈥檚 Optics One of the many works published by the Greek astronomer Ptolemy (a.d. circa 90鈥�170) was Optics. In this book Ptolemy reported the results of refraction experiments he conducted by observing light passing from air into water. Two of his results of refraction = 8.00; (2) angle of incidence = 20.0, angle of refraction = 15.5. Find the percent error for each of Ptolemy鈥檚 measurements, assuming that the index of refraction of water is 1.33. are as follows: (1) angle of incidence = 10.0, angle

When an object is located \(32 \mathrm{~cm}\) to the left of a lens, the image is formed \(17 \mathrm{~cm}\) to the right of the lens. What is the focal length of the lens?

A friend tells you that when she takes off her eyeglasses and holds them \(21 \mathrm{~cm}\) above a printed page, the image of the print is upright but enlarged to \(1.5\) times its actual size. (a) Are the lenses in the glasses concave or convex? Explain. (b) What is the focal length of your friend's glasses?

A beam of light traveling in water \((n=1.33)\) encounters a boundary with another material. If the critical angle for total internal reflection is \(81^{\circ}\), what is the index of refraction of the other material?
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