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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

Expert verified
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

When performing optical spectroscopy (for example, photoluminescence or Raman spectroscopy), a laser beam is focused on the sample to be investigated by means of a lens having a focal distance \(f\). Assume that the laser beam exits a pupil \(D_{o}\) in diameter that is located at a distance \(d_{\mathrm{o}}\) from the focusing lens. For the case when the image of the exit pupil forms on the sample, calculate a) at what distance \(d_{\mathrm{i}}\) from the lens is the sample located and b) the diameter \(D_{i}\) of the laser spot (image of the exit pupil) on the sample. c) What are the numerical results for: \(f=10.0 \mathrm{~cm},\) \(D_{o}=2.00 \mathrm{~mm}, d_{\mathrm{o}}=1.50 \mathrm{~m} ?\)

Where is the image formed if an object is placed \(25 \mathrm{~cm}\) from the eye of a nearsighted person. What kind of a corrective lens should the person wear? a) Behind the retina. Converging lenses. b) Behind the retina. Diverging lenses. c) In front of the retina. Converging lenses. d) In front of the retina. Diverging lenses.

A microscope has a \(2.0-\mathrm{cm}\) focal length eyepiece and a \(0.80-\mathrm{cm}\) objective lens. For a relaxed normal eye, calculate the position of the object if the distance between the lenses is \(16.2 \mathrm{~cm}\)

A plastic cylinder of length \(3.0 \cdot 10^{1} \mathrm{~cm}\) has its ends ground to convex (from the rod outward) spherical surfaces, each having radius of curvature \(1.0 \cdot 10^{1} \mathrm{~cm}\). A small object is placed \(1.0 \cdot 10^{1} \mathrm{~cm}\) from the left end. How far will the image of the object lie from the right end, if the index of refraction of the plastic is \(1.5 ?\)

What type of lens is a magnifying glass? a) converging d) cylindrical b) diverging e) plain c) spherical
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