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Chapter 32: Problem 45

You have a spherical mirror with a radius of curvature of \(+20.0 \mathrm{~cm}\) (so it is concave facing you). You are looking at an object whose size you want to double in the image, so you can see it better. Where should you put the object? Where will the image be, and will it be real or virtual?

Short Answer

Expert verified
Where is the image located, and is it real or virtual? Answer: The object should be placed at a distance of 20 cm from the concave mirror to double the size of its image. The image is located at -40 cm from the mirror, indicating that it is a virtual image.

Step by step solution

01

Recall the mirror and magnification equations

The mirror equation for spherical mirrors is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Where: - \(f\) is the focal length, - \(d_o\) is the object distance (distance from the object to the mirror), and - \(d_i\) is the image distance (distance from the image to the mirror). The magnification equation is: \[ M = -\frac{d_i}{d_o} \] Where: - \(M\) is the magnification factor. Since we want the image to be twice the size of the object, we set \(M = 2\).
02

Calculate the focal length

The radius of curvature \(R\) is given as +20.0 cm. Recall that for concave mirrors, the focal length is half the radius of curvature. Therefore: \[ f = \frac{R}{2} = \frac{20}{2} = 10 \mathrm{~cm} \]
03

Apply the magnification equation M = 2 and solve for \(d_i\) in terms of \(d_o\)

Substitute M = 2 into the magnification equation and solve for \(d_i\): \[ 2 = -\frac{d_i}{d_o} \] \[ d_i = -2d_o \]
04

Substitute \(d_i\) into the mirror equation and solve for \(d_o\)

Replace \(d_i\) with \(-2d_o\) in the mirror equation: \[ \frac{1}{10} = \frac{1}{d_o} - \frac{1}{2d_o} \] Now, solve for \(d_o\): \[ \frac{1}{10} = \frac{1 - 1/2}{d_o} = \frac{1/2}{d_o} \] \[ d_o = 2 \times 10 = 20 \mathrm{~cm} \]
05

Use \(d_o\) to find the image distance \(d_i\)

Now, substitute \(d_o = 20\) cm back into the expression for \(d_i\) from Step 3: \[ d_i = -2d_o = -2(20) = -40 \mathrm{~cm} \]
06

Determine if the image is real or virtual

If the image distance \(d_i\) is positive, the image is real. If the image distance \(d_i\) is negative, the image is virtual. In this case, \(d_i = -40\) cm, so the image is virtual. #Conclusion#: To double the size of the image, place the object at a distance of 20 cm from the concave mirror. The image will be located at -40 cm from the mirror, indicating that it is a virtual image.

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