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Chapter 24: Problem 19

A slide projector has a lens of focal length \(12 \mathrm{cm} .\) Each slide is \(24 \mathrm{mm}\) by \(36 \mathrm{mm}\) (see the figure with Problem 16). The projector is used in a room where the screen is \(5.0 \mathrm{m}\) from the projector. How large must the screen be?

Short Answer

Expert verified
Based on the given information, a slide projector with a lens of focal length 12 cm is used to project slides of size 24 mm x 36 mm (2.4 cm x 3.6 cm) onto a screen 5 meters away. Using lens and magnification formulas, the required screen size to accommodate the projected image is approximately 10.94 cm x 16.4 cm.

Step by step solution

01

List the given information

Focal length of the lens (f) = 12 cm Size of each slide (h1) = 24 mm × 36 mm = 2.4 cm × 3.6 cm Distance between projector and screen (u) = -5 m = -500 cm (note the negative sign, since the object lies on the left side of the lens) We are supposed to find the size of the screen, which will give us the height of the image (h2).
02

Use the lens formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) as follows: 1/f = 1/u + 1/v We need to find the image distance (v). Rearrange the formula to find v: 1/v = 1/f - 1/u Plug in the given values (f and u): 1/v = 1/12 - 1/(-500) Calculate the value of v: 1/v = 1/12 + 1/500 = 629/(12*500) = 629/6000 Now, find the value of v: v = 6000/629 ≈ 9.54 cm
03

Find the magnification (M)

Use the magnification formula to find the magnification (M): M = v/u = -h2/h1 We have the value of v and u. Now, we have to find the size of the image (h2). Rearrange the formula to find h2: h2 = -M * h1 Plug in the values of M and h1: For the vertical size (2.4 cm), h2_vertical = -(9.54/(-500)) * 2.4 ≈ 0.0456 * 2.4 ≈ 0.1094 m ≈ 10.94 cm For the horizontal size (3.6 cm), h2_horizontal = -(9.54/(-500)) * 3.6 ≈ 0.0456 * 3.6 ≈ 0.164 m ≈ 16.4 cm
04

Find the size of the screen

So, the size of the screen must be approximately 10.94 cm × 16.4 cm to accommodate the projected image.

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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.
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