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Chapter 33: Problem 90

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?

Short Answer

Expert verified
Answer: The object will be focused at a distance of \(5.0~cm\) behind the diverging lens.

Step by step solution

01

Find the image position of object due to the converging lens

Since the object is at infinity, the image will be at the focal point of the converging lens. So, the image position due to the converging lens (\(d_{i1}\)) is \(20.0~cm\) behind it.
02

Find the object position for the diverging lens

The object for the diverging lens will be the image formed by the converging lens. We know that the distance between the two lenses is \(15.0~cm\), so the object is \(20.0 - 15.0 = 5.0~cm\) away from the diverging lens. We can denote this distance as \(d_{o2}\).
03

Use Lensmaker's equation for the diverging lens

Lensmaker's equation is given as follows: \(\frac{1}{f} = \frac{1}{d_o} - \frac{1}{d_i}\) For the diverging lens, \(f = -30.0~cm\) and \(d_{o2} = 5.0~cm\). We will now use this equation to find the image position for the diverging lens (\(d_{i2}\)). \(\frac{1}{-30.0} = \frac{1}{5.0} - \frac{1}{d_{i2}}\)
04

Solve for \(d_{i2}\)

We have the equation: \(\frac{1}{-30.0} = \frac{1}{5.0} - \frac{1}{d_{i2}}\) To solve for \(d_{i2}\), we will first subtract \(\frac{1}{5.0}\) from both sides: \(\frac{1}{-30.0} - \frac{1}{5.0} = -\frac{1}{d_{i2}}\) Now we need to find a common denominator for the fractions on the left-hand side of the equation. The common denominator will be \(30.0\): \(\frac{-5-1}{30.0} = -\frac{1}{d_{i2}}\) \(\frac{-6}{30.0} = -\frac{1}{d_{i2}}\) Now we have fractions with the same numerator, so we can easily find \(d_{i2}\): \(d_{i2} = 30.0 \cdot \frac{1}{6} = 5.0~cm\)
05

Final Answer

The final image position is \(5.0~cm\) behind the diverging lens.

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Most popular questions from this chapter

Astronomers sometimes place filters in the path of light as it passes through their telescopes and optical equipment. The filters allow only a single color to pass through. What are the advantages of this? What are the disadvantages?

An unknown lens forms an image of an object that is \(24 \mathrm{~cm}\) away from the lens, inverted, and a factor of 4 larger in size than the object. Where is the object located? a) \(6 \mathrm{~cm}\) from the lens on the same side of the lens b) \(6 \mathrm{~cm}\) from the lens on the other side of the lens c) \(96 \mathrm{~cm}\) from the lens on the same side of the lens d) \(96 \mathrm{~cm}\) from the lens on the other side of the lens e) No object could have formed this image.

In H.G. Wells's classic story The Invisible Man, a man manages to change the index of refraction of his body to 1.0 ; thus, light would not bend as it enters his body (assuming he is in air and not swimming). If the index of refraction of his eyes were equal to one, would he be able to see? If so, how would things appear?

Where is the image formed if an object is placed \(25 \mathrm{~cm}\) from the eye of a nearsighted person. What kind of a corrective lens should the person wear? a) Behind the retina. Converging lenses. b) Behind the retina. Diverging lenses. c) In front of the retina. Converging lenses. d) In front of the retina. Diverging lenses.

A converging lens will be used as a magnifying glass. In order for this to work, the object must be placed at a distance a) \(d_{\mathrm{o}}>f\). c) \(d_{\mathrm{o}}
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