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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

Two distant stars are separated by an angle of 35 arcseconds. If you have a refracting telescope whose objective lens focal length is \(3.5 \mathrm{~m}\), what focal length eyepiece do you need in order to observe the stars as though they were separated by 35 arcminutes?

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The object (upright arrow) in the following system has a height of \(2.5 \mathrm{~cm}\) and is placed \(5.0 \mathrm{~cm}\) away from a converging (convex) lens with a focal length of \(3.0 \mathrm{~cm}\). What is the magnification of the image? Is the image upright or inverted? Confirm your answers by ray tracing.

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