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Chapter 2: Problem 100

Unreasonable Results Your friends show you an image through a microscope. They tell you that the microscope has an objective with a 0.500 -cm focal length and an eyepiece with a \(5.00-\mathrm{cm}\) focal length. The resulting overall magnification is 250,000 . Are these viable values for a microscope? Unless otherwise stated, the lens-to-retina distance is 2.00 \(\mathrm{cm}\)

Short Answer

Expert verified
The given microscope configuration has an objective lens with a 0.500cm focal length and an eyepiece with a 5.00cm focal length, which results in an overall magnification of 7, as calculated using the appropriate equations. However, the given overall magnification is 250,000. Since the calculated magnification is significantly smaller than the given magnification, the given values are not viable for a microscope configuration.

Step by step solution

01

Calculate the objective lens magnification

First, we will find the magnification produced by the objective lens using the equation: \(M_1 = 1 + \frac{f_1}{d_1}\) where \(M_1\) refers to the magnification of objective, \(f_1\) is the focal length of the objective lens, and \(d_1\) corresponds to the distance from the lens to the object being viewed. For a practical microscope, \(d_1\) is nearly equal to the focal length of the objective lens. So, we'll consider \(d_1 = 0.500cm\). Now, calculate the magnification, \(M_1\): \(M_1 = 1 + \frac{0.500}{0.500} = 1 + 1 = 2\) The magnification by the objective lens is 2.
02

Calculate the eyepiece lens magnification

Next, we will find the magnification produced by the eyepiece lens using the equation: \(M_2 = 1 + \frac{f_2}{d_2}\) where \(M_2\) refers to the magnification of the eyepiece, \(f_2\) is the focal length of the eyepiece lens, and \(d_2\) corresponds to the distance from the lens to the observer's retina. The problem states the lens-to-retina distance, \(d_2\), is 2.00 cm. Given the focal length of the eyepiece, \(f_2 = 5.00 cm\), we can calculate the magnification \(M_2\): \(M_2 = 1 + \frac{5.00}{2.00} = 1 + 2.50 = 3.5\) The magnification by the eyepiece lens is 3.5.
03

Calculate the overall magnification

Now, we will calculate the total magnification produced by both lenses combined, using the formula: Overall Magnification = \(M_1 \times M_2\) We found \(M_1 = 2\) and \(M_2 = 3.5\), so the overall magnification is: Overall Magnification = \(2 \times 3.5 = 7\) The calculated overall magnification for the given microscope configuration is 7.
04

Compare calculated overall magnification with given magnification

We calculated the overall magnification to be 7, whereas the given overall magnification is 250,000. Since the calculated magnification is much smaller than the given magnification, we can conclude that the given values are not viable for a microscope configuration.

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Most popular questions from this chapter

Unless otherwise stated, the lens-to-retina distance is 2.00 Calculate the power of the eye when viewing an object \(3.00 \mathrm{m}\) away.

What is the angular size of the Moon if viewed from a binocular that has a focal length of \(1.2 \mathrm{cm}\) for the eyepiece and a focal length of \(8 \mathrm{cm}\) for the objective? Use the radius of the moon \(1.74 \times 10^{6} \mathrm{m}\) and the distance of the moon from the observer to be \(3.8 \times 10^{8} \mathrm{m}\)

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An object of height 2 \(\mathrm{cm}\) is placed at \(50 \mathrm{cm}\) in front of a diverging lens of focal length \(40 \mathrm{cm} .\) Behind the lens, there is a convex mirror of focal length \(15 \mathrm{cm}\) placed 30 cm from the converging lens. Find the location, orientation, and size of the final image.

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