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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

A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)

A refracting telescope is \(45.0 \mathrm{cm}\) long and the caption states that the telescope magnifies images by a factor of \(30.0 .\) Assuming these numbers are for viewing an object an infinite distance away with minimum eyestrain, what is the focal length of each of the two lenses?

The uncorrected far point of Colin's eye is \(2.0 \mathrm{m} .\) What refractive power contact lens enables him to clearly distinguish objects at large distances?
A refracting telescope is used to view the Moon. The focal lengths of the objective and eyepiece are \(+2.40 \mathrm{m}\) and \(+16.0 \mathrm{cm},\) respectively. (a) What should be the distance between the lenses? (b) What is the diameter of the image produced by the objective? (c) What is the angular magnification?
A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.
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