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Chapter 24: Problem 55

Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find the intermediate and final image distances relative to the corresponding lenses. (b) What is the total magnification? (c) What is the height of the final image?

Short Answer

Expert verified
Answer: The intermediate and final image distances relative to the corresponding lenses are 60 cm and 60 cm respectively. The total magnification is 18, and the height of the final image is 54 cm.

Step by step solution

01

Find the intermediate image distance

Using the thin lens formula, we can find the image distance for the first lens: \(1/f = 1/d_o + 1/d_i\) Where \(d_o\) is the object distance, \(d_i\) is the image distance, and \(f\) is the focal length. Let's plug in the known values for the first lens and solve for the intermediate image distance, \(d_{i1}\): \(1/15 = 1/20 + 1/d_{i1}\) Now, solve for \(d_{i1}\): \(d_{i1} = 60\,\text{cm}\)
02

Image produced by first lens becomes object for second lens

The image produced by the first lens becomes the object for the second lens. We can find the object distance (\(d_{o2}\)) for the second lens by subtracting the distance between the lenses from the image distance \(d_{i1}\). \(d_{o2} = d_{i1} - 50\) \(d_{o2} = 60 - 50\) \(d_{o2} = 10\,\mathrm{cm}\)
03

Find the final image distance for the second lens

Now, let's use the lens formula to find the final image distance, \(d_{i2}\), for the second lens: \(1/f = 1/d_o + 1/d_i\) Plug in the known values for the second lens: \(1/12 = 1/10 + 1/d_{i2}\) Solve for \(d_{i2}\): \(d_{i2} = 60\,\mathrm{cm}\)
04

Find the magnification for each lens

The magnification of each lens can be found using the formula: \(m = -d_i / d_o\) Calculate the magnification for the first lens, \(m_1\): \(m_1 = -d_{i1} / 20\) \(m_1 = -60 / 20\) \(m_1 = -3\) Calculate the magnification for the second lens, \(m_2\): \(m_2 = -d_{i2} / 10\) \(m_2 = -60 / 10\) \(m_2 = -6\)
05

Find the total magnification

To find the total magnification, multiply the magnification of each lens: \(M = m_1 \times m_2\) \(M = (-3) \times (-6)\) \(M = 18\)
06

Find the final image height

To find the height of the final image, multiply the total magnification with the object's height: \(h_{i} = M \times h_o\) \(h_{i} = 18 \times 3\) \(h_{i} = 54\,\mathrm{cm}\) #Summary# The intermediate and final image distances relative to the corresponding lenses are \(d_{i1} = 60\,\mathrm{cm}\) and \(d_{i2} = 60\,\mathrm{cm}\), respectively. The total magnification is \(18\), and the height of the final image is \(54\,\mathrm{cm}\).

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Most popular questions from this chapter

You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length $3.70 \mathrm{cm}\( located \)6.00 \mathrm{cm}$ to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
A slide projector, using slides of width \(5.08 \mathrm{cm},\) produces an image that is \(2.00 \mathrm{m}\) wide on a screen \(3.50 \mathrm{m}\) away. What is the focal length of the projector lens?
A dissecting microscope is designed to have a large distance between the object and the objective lens. Suppose the focal length of the objective of a dissecting microscope is \(5.0 \mathrm{cm},\) the focal length of the eyepiece is \(4.0 \mathrm{cm},\) and the distance between the lenses is $32.0 \mathrm{cm} .$ (a) What is the distance between the object and the objective lens? (b) What is the angular magnification?
A man requires reading glasses with \(+2.0 \mathrm{D}\) power to read a book held \(40.0 \mathrm{cm}\) away with a relaxed eye. Assume the glasses are $2.0 \mathrm{cm}$ from his eyes. (a) What is his uncorrected far point? (b) What refractive power lenses should he use for distance vision? (c) His uncorrected near point is \(1.0 \mathrm{m} .\) What should the refractive powers of the two lenses in his bifocals be to give him clear vision from \(25 \mathrm{cm}\) to infinity?
An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
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