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Chapter 24: Problem 9

Cameras 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
Given a two-lens system with a desired image position, a focal length of the first lens, and the position of the object, we calculated the focal length of the second lens to be 3.63 cm, indicating that it is a converging lens. The total magnification of the system is 1.79, and the final image height is 21.5 cm.

Step by step solution

01

Image position due to the first lens

Using the lens equation: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] Where \(f\) is the focal length of the lens, \(d_o\) is the distance to the object, and \(d_i\) is the distance to the image. Given the focal length \(f_1 = 3.70\) cm and distance to the object \(d_{o1} = 6.00\) cm, we can solve for the distance to the image \(d_{i1}\): \[\frac{1}{3.70} = \frac{1}{6.00} + \frac{1}{d_{i1}}\] Solving for \(d_{i1}\), we find that \(d_{i1} = 11.10\) cm.
02

Image position due to the second lens

Now let's find the image position due to the second lens. First, we need to find the object distance for the second lens, which is equal to the distance between the two lenses minus the image distance produced by the first lens. Given the distance between the two lenses is \(24.65 - 6.00 = 18.65\) cm and the image distance produced by the first lens is \(11.10\) cm, the object distance for the second lens is: \(d_{o2} = 18.65 - 11.10 = 7.55\) cm
03

Finding the focal length of the second lens

We can now use the lens equation to find the focal length of the second lens. We know the object distance \(d_{o2} = 7.55\) cm and the image distance \(d_{i2} = 32.0 - 24.65 = 7.35\) cm. Plugging these values into the lens equation: \[\frac{1}{f_2} = \frac{1}{7.55} + \frac{1}{7.35}\] Solving this equation for \(f_2\), we find that the focal length of the second lens is \(f_2 = 3.63\) cm.
04

Determining the lens type

Since the focal length of the second lens is positive, it is a converging lens.
05

Calculating total magnification

To find the total magnification, we first need to find the magnification due to each lens. The magnification is calculated as: \[M = -\frac{d_i}{d_o}\] For the first lens, \(M_1 = -\frac{11.10}{6.00} = -1.85\). For the second lens, \(M_2 = -\frac{7.35}{7.55} = -0.97\). The total magnification is the product of the magnifications of each lens: \[M = M_1 \times M_2 = -1.85 \times -0.97 = 1.79\]
06

Finding the image height

We are given the object height \(h_o = 12.0\) cm. The image height is the product of the total magnification and the object height: \[h_i = M \times h_o = 1.79 \times 12.0 = 21.5 \ cm\] In conclusion, the focal length of the second lens is \(3.63\) cm, the lens is converging, the total magnification is \(1.79\), and the image height is \(21.5\) cm.

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Most popular questions from this chapter

Two converging lenses are placed \(88.0 \mathrm{cm}\) apart. An object is placed \(1.100 \mathrm{m}\) to the left of the first lens. which has a focal length of \(25.0 \mathrm{cm} .\) The final image is located \(15.0 \mathrm{cm}\) to the right of the second lens. (a) What is the focal length of the second lens? (b) What is the total magnification?

A converging lens with a focal length of \(15.0 \mathrm{cm}\) and a diverging lens are placed \(25.0 \mathrm{cm}\) apart, with the converging lens on the left. A 2.00 -cm-high object is placed \(22.0 \mathrm{cm}\) to the left of the converging lens. The final image is \(34.0 \mathrm{cm}\) to the left of the converging lens. (a) What is the focal length of the diverging lens? (b) What is the height of the final image? (c) Is the final image upright or inverted?
Repeat Problem \(40(\mathrm{c})\) using a different eyepiece that gives an angular magnification of 5.00 for a final image at the viewer's near point \((25.0 \mathrm{cm})\) instead of at infinity.
A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
A camera with a 50.0 -mm lens can focus on objects located from $1.5 \mathrm{m}$ to an infinite distance away by adjusting the distance between the lens and the film. When the focus is changed from that for a distant mountain range to that for a flower bed at \(1.5 \mathrm{m},\) how far does the lens move with respect to the film?
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