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Chapter 33: Problem 28

An object of height \(h\) is placed at a distance \(d_{0}\) on the left side of a converging lens of focal length \(f\left(f

Short Answer

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Question: An object is placed in front of a converging lens, and an image is formed 3 times the focal length, f, to the right of the lens. a) Calculate the distance the object needs to be placed on the left side of the lens, d₀. b) Calculate the magnification, M, of the image. Answer: a) The object must be placed at a distance of \(\frac{3f}{2}\) on the left side of the lens. b) The magnification of the image is -2, meaning the image is flipped and twice the size of the object.

Step by step solution

01

Solving for \(d_0\) with the lens equation

We have the lens equation: \(\frac{1}{f} = \frac{1}{d_0} + \frac{1}{d}\). Given that \(d = 3f\), we can substitute this into the lens equation and solve for \(d_0\): \(\frac{1}{f} = \frac{1}{d_0} + \frac{1}{3f}\) Now, we'll solve for \(d_0\): \(\frac{1}{d_0} =\frac{1}{f} - \frac{1}{3f} = \frac{3-1}{3f} = \frac{2}{3f}\) Therefore, \(d_0 = \frac{3f}{2}\) So the object must be placed at a distance \(\frac{3f}{2}\) on the left side of the lens for the image to form at a distance \(3f\) on the right side of the lens.
02

Calculating the magnification

Now that we have found \(d_0 = \frac{3f}{2}\), we can use the formula for magnification, \(M = \frac{-d}{d_0}\): \(M = \frac{-3f}{\frac{3f}{2}}\) Multiplying by \(\frac{2}{1}\) to clear the fraction: \(M = \frac{-2(3f)}{1(3f)}\) Canceling the common factor of \(3f\) in the numerator and denominator: \(M = -2\) The magnification is -2, indicating that the image is flipped (which is denoted by the negative sign) and is twice the size of the object.

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Most popular questions from this chapter

You have found in the lab an old microscope, which has lost its eyepiece. It still has its objective lens, and markings indicate that its focal length is \(7.00 \mathrm{~mm}\). You can put in a new eyepiece, which goes in \(20.0 \mathrm{~cm}\) from the objective. You need a magnification of about 200. Assume you want the comfortable viewing distance for the final image to be \(25.0 \mathrm{~cm}\). You find in a drawer eyepieces marked \(2.00-, 4.00-,\) and \(8.00-\mathrm{cm}\) focal length. Which is your best choice?

What kind of lens is used in eyeglasses to correct the vision of someone who is a) nearsighted? b) farsighted?

Astronomers sometimes place filters in the path of light as it passes through their telescopes and optical equipment. The filters allow only a single color to pass through. What are the advantages of this? What are the disadvantages?

Jack has a near point of \(32 \mathrm{~cm}\) and uses a magnifier of 25 diopter. a) What is the magnification if the final image is at infinity? b) What is the magnification if the final image is at the near point?

An object is \(6.0 \mathrm{~cm}\) from a thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm},\) determine the image distance.
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