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Chapter 24: Problem 9

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?

Short Answer

Expert verified
(a) The focal length of the second lens is approximately 5.27 cm. (b) The second lens is converging. (c) The total magnification is approximately 0.677. (d) The image height is approximately 8.12 cm.

Step by step solution

01

Apply the first lens equation

In order to find the position of the image formed by the first lens, we can use the lens equation: $$\frac{1}{f_{1}} = \frac{1}{d_{o1}} + \frac{1}{d_{i1}}$$ where \(f_{1}\) is the focal length of the first lens, \(d_{o1}\) is the object distance for the first lens, \(d_{i1}\) is the image distance formed by the first lens. By substituting the given values, we can find the \(d_{i1}\): $$\frac{1}{3.7} = \frac{1}{6} + \frac{1}{d_{i1}}$$
02

Find the image distance for the first lens

Next, we will solve the equation obtained in Step 1 for \(d_{i1}\): $$d_{i1} = \frac{1}{(\frac{1}{3.7} - \frac{1}{6})} \approx 6.267 \mathrm{cm}$$
03

Find the object distance for the second lens

Since the image formed by the first lens will act as a virtual object for the second lens, we can find the object distance for the second lens (\(d_{o2}\)): $$d_{o2} = 24.65 - 6 = 18.65 \mathrm{cm}$$
04

Use the given image distance for the combined lenses

The combined lenses are given to produce an image at \(32.0 \mathrm{cm}\) to the right of the object. To find the combined image distance for the second lens, we can use the following formula: $$d_{i2} = 32 - 24.65 = 7.35 \mathrm{cm}$$
05

Apply the second lens equation

For the second lens, we can use the lens equation to find the focal length \(f_{2}\): $$\frac{1}{f_{2}} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}}$$ By substituting the values of \(d_{o2}\) and \(d_{i2}\), we can solve for the focal length of the second lens, \(f_{2}\).
06

Find the focal length of the second lens

Solving the equation obtained in Step 5 for \(f_{2}\), we get: $$f_{2} = \frac{1}{(\frac{1}{18.65} + \frac{1}{7.35})} \approx 5.27 \mathrm{cm}$$ Now we can answer part (a) of the problem: (a) The focal length of the second lens is approximately \(5.27 \mathrm{cm}\).
07

Determine if the second lens is converging or diverging

Since the focal length of the second lens is positive, it is a converging lens: (b) The second lens is converging.
08

Find the total magnification

The total magnification is the product of the magnifications of both lenses: $$M = M_{1} \cdot M_{2} = \left(\frac{-d_{i1}}{d_{o1}}\right) \cdot \left(\frac{-d_{i2}}{d_{o2}}\right)$$ Plugging in the values, we get: $$M = \left(\frac{-6.267}{6}\right) \cdot \left(\frac{-7.35}{18.65}\right) \approx 0.677$$ (c) The total magnification is approximately 0.677.
09

Find the image height

The image height can be found by multiplying the total magnification by the object height: $$h_{i} = M \cdot h_{o} = 0.677 \cdot 12.0 \mathrm{cm} \approx 8.12 \mathrm{cm}$$ (d) The image height is approximately \(8.12 \mathrm{cm}\).

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Most popular questions from this chapter

Suppose that the lens system (cornea + lens) in a particular eye has a focal length that can vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm},\) but the distance from the lens system to the retina is only \(1.90 \mathrm{cm} .\) (a) Is this eye nearsighted or farsighted? Explain. (b) What range of distances can the eye see clearly without corrective lenses?
A refracting telescope has an objective lens with a focal length of $2.20 \mathrm{m}\( and an eyepiece with a focal length of \)1.5 \mathrm{cm} .$ If you look through this telescope the wrong way, that is, with your eye placed at the objective lens, by what factor is the angular size of an observed object reduced?
A converging lens and a diverging lens, separated by a distance of $30.0 \mathrm{cm},$ are used in combination. The converging lens has a focal length of \(15.0 \mathrm{cm} .\) The diverging lens is of unknown focal length. An object is placed \(20.0 \mathrm{cm}\) in front of the converging lens; the final image is virtual and is formed \(12.0 \mathrm{cm}\) before the diverging lens. What is the focal length of the diverging lens?
A cub scout makes a simple microscope by placing two converging lenses of +18 D at opposite ends of a \(28-\mathrm{cm}^{-}\) long tube. (a) What is the tube length of the microscope? (b) What is the angular magnification? (c) How far should an object be placed from the objective lens?
Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$
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