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Chapter 33: Problem 88

Jack has a near point of \(32 \mathrm{~cm}\) and uses a magnifier of 25 diopter. a) What is the magnification if the final image is at infinity? b) What is the magnification if the final image is at the near point?

Short Answer

Expert verified
Answer: The magnification when the final image is at infinity is 9, and the magnification when the final image is at the near point is 8.

Step by step solution

01

Calculate the focal length of the magnifier

First, we need to find the focal length of the magnifier. We can do this using the lens power formula: \(f = \frac{1}{P}\) Where \(f\) represents the focal length, and \(P\) represents the power of the lens in diopters. Substitute the given value for P: \(f = \frac{1}{25}\) So we have: \(f = 0.04 \mathrm{~m}\) or \(4 \mathrm{~cm}\) The focal length of the magnifier is 4 cm.
02

Determine the image position when the final image is at infinity

When the image is at infinity, the object is at the focal point of the magnifying lens. Therefore, the object distance \(u\) is equal to the focal length \(f\). In this case, we have: \(u = f = 4 \mathrm{~cm}\)
03

Calculate the first magnification: when the final image is at infinity

We can calculate the magnification by using the formula: \(M = 1 + \frac{d_{n}}{f}\) Where \(M\) represents the magnification, \(d_{n}\) represents the near point distance, and \(f\) is the focal length of the lens. Plug in the values for \(d_{n}\) and \(f\): \(M = 1 + \frac{32 \mathrm{~cm}}{4 \mathrm{~cm}}\) So, we have: \(M = 1 + 8 = 9\) The magnification when the final image is at infinity is 9.
04

Determine the image position when the final image is at the near point

When the final image is at the near point, the image distance \(v\) is equal to the near point distance \(d_{n}\) (32 cm). To find the object distance (\(u\)) when the final image is at the near point, we can use the lens formula: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) Substitute the values for \(f\) and \(v\): \(\frac{1}{4 \mathrm{~cm}} = \frac{1}{u} + \frac{1}{32 \mathrm{~cm}}\)
05

Solve for the object distance \(u\) for the near point case

Now solve for \(u\) in the lens formula: \(\frac{1}{u} = \frac{1}{4 \mathrm{~cm}} - \frac{1}{32 \mathrm{~cm}}\) u = 4.57 cm The object distance when the final image is at the near point is 4.57 cm.
06

Calculate the second magnification when the final image is at the near point

Next, we can calculate the magnification for the near point case using the same magnification formula: \(M = 1 + \frac{d_{n}}{f}\) Using the object distance for the near point case: \(M = 1 + \frac{32 \mathrm{~cm}}{4.57 \mathrm{~cm}}\) M = 8 The magnification when the final image is at the near point is 8. To summarize the results: a) The magnification when the final image is at infinity is 9. b) The magnification when the final image is at the near point is 8.

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Most popular questions from this chapter

An instructor wants to use a lens to project a real image of a light bulb onto a screen \(1.71 \mathrm{~m}\) from the bulb. In order to get the image to be twice as large as the bulb, what focal length lens will be needed?

What kind of lens is used in eyeglasses to correct the vision of someone who is a) nearsighted? b) farsighted?

For a microscope to work as intended, the separation between the objective lens and the eyepiece must be such that the intermediate image produced by the objective lens will occur at a distance (as measured from the optical center of the eyepiece) a) slightly larger than the focal length. b) slightly smaller than the focal length. c) equal to the focal length. d) The position of the intermediate image is irrelevant.

A camera has a lens with a focal length of \(60 . \mathrm{mm} .\) Suppose you replace the normal lens with a zoom lens whose focal length can be varied from \(35 . \mathrm{mm}\) to \(250 . \mathrm{mm}\) and use the camera to photograph an object at infinity. Compared to a 60.-mm lens, what magnification of the image would be achieved using the \(240 .-\mathrm{mm}\) focal length?

A diverging lens with \(f=-30.0 \mathrm{~cm}\) is placed \(15.0 \mathrm{~cm}\) behind a converging lens with \(f=20.0 \mathrm{~cm}\). Where will an object at infinity in front of the converging lens be focused?
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