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Chapter 23: Problem 69

A standard \(\cdots 35-\mathrm{mm}^{\prime \prime}\) slide measures $24.0 \mathrm{mm}\( by \)36.0 \mathrm{mm} .$ Suppose a slide projector produces a \(60.0-\mathrm{cm}\) by 90.0 -cm image of the slide on a screen. The focal length of the lens is \(12.0 \mathrm{cm} .\) (a) What is the distance between the slide and the screen? (b) If the screen is moved farther from the projector, should the lens be moved closer to the slide or farther away?

Short Answer

Expert verified
Answer: The distance between the slide and the screen is 312 cm. If the screen moves further away, the lens should be moved closer to the slide to maintain the same magnification and keep the image clear.

Step by step solution

01

Understand the given parameters

: We are given the following information: 1. Slide dimensions: \(24 \ mm \ \text{x} \ 36 \ mm\) 2. Image dimensions: \(60 \ cm \ \text{x} \ 90 \ cm\) 3. Focal length of the lens: \(12 \ cm\) Note: To keep the units consistent, let's convert the slide dimensions into centimeters as 1m = 100cm. Slide dimensions: \(2.4 \ cm \ \text{x} \ 3.6 \ cm\)
02

Calculate the magnification

: For this problem, we can determine the magnification by comparing the height of the image with the height of the slide. Magnification, \(M = \dfrac{height \ of \ image}{height \ of \ slide}=\dfrac{60 \ cm}{2.4 \ cm}=25\)
03

Use the lens formula

: The lens formula is: \(\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}\) Where: - \(f\) is the focal length - \(d_o\) is the object distance (distance between slide and lens) - \(d_i\) is the image distance (distance between lens and screen) We already know the focal length \(f\), and we can find \(d_o\) or \(d_i\) using the magnification formula: \(M=\dfrac{d_i}{d_o}\) We can solve these equations to find the distances \(d_o\) and \(d_i\).
04

Find the object distance (d_o)

: We can use the magnification formula to find the object distance: \(d_o=\dfrac{d_i}{M}\) Since we don't yet have the value for \(d_i\), let's write \(d_o\) in terms of \(d_i\): \(d_o=\dfrac{1}{M}d_i\)
05

Find the image distance (d_i) using the lens formula

: Now, let's use the lens formula and replace \(d_o\) with \(\dfrac{1}{M}d_i\): \(\dfrac{1}{f} = \dfrac{1}{\dfrac{1}{M}d_i} + \dfrac{1}{d_i}\) Plug in the values (\(M = 25\) and \(f = 12 \ cm\)): \(\dfrac{1}{12} = \dfrac{1}{\dfrac{1}{25}d_i} + \dfrac{1}{d_i}\) Now, let's solve for \(d_i\): \(d_i = 300 \ cm\)
06

Calculate the distance between the slide and the screen

: Now that we've found the image distance (\(d_i\)), we can find the object distance (\(d_o\)): \(d_o = \dfrac{1}{M} d_i = \dfrac{1}{25} (300 \ cm) = 12 \ cm\) Finally, let's find the total distance between the slide and the screen, which is the sum of \(d_o\) and \(d_i\): Total distance \(= d_o + d_i = 12 \ cm + 300 \ cm = 312 \ cm\) So, the distance between the slide and the screen is \(312 \ cm\).
07

Adjusting the lens for a farther screen

: If the screen is moved farther from the projector, the image distance (\(d_i\)) increases. To maintain the same magnification and keep the image clear, the lens must also be moved according to the lens formula. As \(d_i\) increases, in order to maintain the same focal length, \(d_o\) must decrease. So, the lens should be moved closer to the slide if the screen is moved farther away.

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