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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

Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?
The distance from the lens system (cornea + lens) of a particular eye to the retina is \(1.75 \mathrm{cm} .\) What is the focal length of the lens system when the eye produces a clear image of an object \(25.0 \mathrm{cm}\) away?
An insect that is 5.00 mm long is placed \(10.0 \mathrm{cm}\) from a converging lens with a focal length of \(12.0 \mathrm{cm} .\) (a) What is the position of the image? (b) What is the size of the image? (c) Is the image upright or inverted? (d) Is the image real or virtual? (e) What is the angular magnification if the lens is close to the eye?
Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?
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