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Chapter 24: Problem 58

The Ortiz family is viewing slides from their summer vacation trip to the Grand Canyon. Their slide projector has a projection lens of \(10.0-\mathrm{cm}\) focal length and the screen is located \(2.5 \mathrm{m}\) from the projector. (a) What is the distance between the slide and the projection lens? (b) What is the magnification of the image? (c) How wide is the image of a slide of width \(36 \mathrm{mm}\) on the screen? (See the figure with Problem 16 .)

Short Answer

Expert verified
Question: A slide projector has a projection lens of focal length 10 cm. It is used to project a slide of width 3.6 cm on a screen 2.5 m away. Determine the distance between the slide and the projection lens, the magnification of the image, and the width of the image on the screen. Answer: The distance between the slide and the projection lens is 10.42 cm, the magnification of the image is 24, and the width of the image on the screen is 86.4 cm.

Step by step solution

01

Variables and formulas

Given: Focal length, f = 10 cm Screen distance, D = 2.5 m (Converting to cm, D = 250 cm) Slide width, w = 36 mm (Converting to cm, w = 3.6 cm) We need to find: (a) Distance between the slide and the projection lens, d_object (b) Magnification of the image, M (c) Width of the image on the screen, W We will use the following formulas: Thin lens equation: \(\frac{1}{f} = \frac{1}{d_{object}} + \frac{1}{d_{image}}\) Magnification formula: \(M = \frac{d_{image}}{d_{object}}\)
02

Find distance between slide and projection lens (d_object)

Using the thin lens equation, we need to find the distance between the slide (object) and the projection lens (d_object). \(\frac{1}{f} = \frac{1}{d_{object}} + \frac{1}{d_{image}}\) We know the focal length (f) and screen distance (d_image). Now, we can solve for d_object. \(\frac{1}{d_{object}} = \frac{1}{f} - \frac{1}{d_{image}}\) Substitute the given values: \(\frac{1}{d_{object}} = \frac{1}{10} - \frac{1}{250}\) Now, we can find d_object. \(d_{object} = \frac{1}{\frac{1}{10} - \frac{1}{250}} = \frac{2500}{240} = 10.42 cm\) So, the distance between the slide and the projection lens is 10.42 cm.
03

Find magnification of the image (M)

Using the magnification formula, we can find the magnification of the image. \(M = \frac{d_{image}}{d_{object}}\) Substitute the values: \(M = \frac{250}{10.42} = 24\) The magnification of the image is 24.
04

Find the width of the image on the screen (W)

Now, we can find the width of the image on the screen using the magnification and slide width. \(W = M \times w\) Substitute the values: \(W = 24 \times 3.6 = 86.4 cm\) So, the width of the image on the screen is 86.4 cm.

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Most popular questions from this chapter

The wing of an insect is \(1.0 \mathrm{mm}\) long. When viewed through a microscope, the image is \(1.0 \mathrm{m}\) long and is located \(5.0 \mathrm{m}\) away. Determine the angular magnification.

A refracting telescope is \(45.0 \mathrm{cm}\) long and the caption states that the telescope magnifies images by a factor of \(30.0 .\) Assuming these numbers are for viewing an object an infinite distance away with minimum eyestrain, what is the focal length of each of the two lenses?
(a) What is the angular size of the Moon as viewed from Earth's surface? See the inside back cover for necessary information. (b) The objective and eyepiece of a refracting telescope have focal lengths \(80 \mathrm{cm}\) and \(2.0 \mathrm{cm}\) respectively. What is the angular size of the Moon as viewed through this telescope?
A microscope has an eyepiece of focal length \(2.00 \mathrm{cm}\) and an objective of focal length \(3.00 \mathrm{cm} .\) The eyepiece produces a virtual image at the viewer's near point \((25.0 \mathrm{cm}\) from the eye). (a) How far from the eyepiece is the image formed by the objective? (b) If the lenses are \(20.0 \mathrm{cm}\) apart, what is the distance from the objective lens to the object being viewed? (c) What is the angular magnification?
A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.
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