Physics 107 HOMEWORK ASSIGNMENT #20
Cutnell & Johnson, 7th edition
Chapter 26: Problems 52, 72, 74, 104, 114
*52 Concept Simulation 26.4 provides the option of exploring the ray diagram that applies to this problem. The distance between an object and its image formed by a diverging lens is 49.0 cm. The focal length of the lens is −233.0 cm. Find (a) the image distance and (b) the object distance.
72 A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.660 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes?
**74 The contacts worn by a farsighted person allow her to see objects clearly that are as close as 25.0 cm, even though her uncorrected near point is 79.0 cm from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of 217 cm from her eyes. (a) How far away is the poster actually located? (b) If the poster is 0.350 m tall, how tall is the image formed by the contacts?
104 A slide projector has a converging lens whose focal length is 105.00 mm. (a) How far (in meters) from the lens must the screen be located if a slide is placed 108.00 mm from the lens? (b) If the slide measures , what are the dimensions (in mm) of its image?
*114 An object is in front of a converging lens (f = 0.30 m). The magnification of the lens is m = 4.0. (a) Relative to the lens, in what direction should the object be moved so that the magnification changes to m = −4.0? (b) Through what distance should the object be moved?
52. REASONING A diverging lens always produces a virtual image, so that the image distance di is negative. Moreover, the object distance do is positive. Therefore, the distance between the object and the image is do+di=49.0cm, rather than do–di=49.0cm. The equation do+di=49.0cm and the thin-lens equation constitute two equations in two unknowns, and we will solve them simultaneously to obtain values for di and do.
SOLUTION
a. Solving the equation do+di=49.0cm for do, substituting the result into the thin-lens equation, and suppressing the units give
/ (1)Grouping the terms on the left of Equation (1) over a common denominator, we have
/ (2)Cross-multiplying and rearranging in Equation (2) gives
/ (3)Using the quadratic formula to solve Equation (3), we obtain
We have discarded the positive root, because we know that di must be negative for the virtual image.
b. Using the fact that do+di=49.0cm, we find that the object distance is
72. REASONING The eyeglasses form a virtual image of the newspaper page at her near point. The image distance is negative, and we can calculate it by using the thin-lens equation. In doing so, we need to keep two things in mind. First, we must take into account the fact that she wears her eyeglasses 2.00cm from her eyes. This is because the distances that appear in the thin-lens equation are measured with respect to the lens of the eyeglasses, not with respect to her eyes. Thus, the object distance (the location of the newspaper page) is do=42.00cm2.00cm. Second, in using the thin-lens equation we must have a value for the focal length f, and this is not directly given. However, we do know the refractive power of the lens, which is the reciprocal of the focal length, according to Equation26.8. After calculating the image distance, we can obtain her near point with respect to the eyeglasses by taking the magnitude of this negative number. Finally, we will account for the fact that the eyeglasses are 2.00cm from her eyes.
SOLUTION According to the thin-lens equation, the reciprocal of the image distance di is
To use this expression, we need a value for the focal length, and we use Equation26.8 for the refractive power to obtain it. This equation indicates that the refractive power in diopters is the reciprocal of the focal length in meters, so we have
Remembering that the eyeglasses are worn 2.00cm from the eyes, we can now apply the thin-lens equation:
The magnitude of this value for di is 119cm and gives the location of the woman’s near point with respect to the eyeglasses. Since the eyeglasses are worn 2.00cm from the eyes, the near point is located at a distance of 121cm from the eyes.
74. REASONING AND SOLUTION If the near point is 79.0 cm, then di = – 79.0 cm, and do=25.0 cm. Using the thin-lens equation, we find that the focal length of the correcting lens is
a. The distance to the poster can be obtained as follows:
or =
b. The image size is
104. REASONING The distance from the lens to the screen, the image distance, can be obtained directly from the thin-lens equation, Equation 26.6, since the object distance and focal length are known. The width and height of the image on the screen can be determined by using Equation 26.7, the magnification equation.
SOLUTION
a. The distance di to the screen is
so that .
b. According to the magnification equation, the width and height of the image on the screen are
Width /The width is .
Height /
The height is .
114. REASONING To find the distance through which the object must be moved, we must obtain the object distances for the two situations described in the problem. To do this, we combine the thin-lens equation and the magnification equation, since data for the magnification is given.
SOLUTION
a. Since the magnification is positive, the image is upright, and the object must be located within the focal point of the lens, as in Figure 26.28. When the magnification is negative and has a magnitude greater than one, the object must be located between the focal point and the point that is at a distance of twice the focal length from the lens, as in Figure 26.27. Therefore, the object should be moved .
b. According to the thin-lens equation, we have
/ (1)According to the magnification equation, with di expressed as in Equation (2), we have
/ (2)Applying Equation (2) to the two cases described in the problem, we have
/ (3)/ (4)
Subtracting Equation (3) from Equation (4), we find that the object must be moved away from the lens by an additional distance of