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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

Mirrors for astronomical instruments are invariably first-surface mirrors: The reflective coating is applied on the surface exposed to the incoming light. Household mirrors, on the other hand, are second-surface mirrors: The coating is applied to the back of the glass or plastic material of the mirror. (You can tell the difference by bringing the tip of an object close to the surface of the mirror. Object and image will nearly touch with a first-surface mirror; a gap will remain between them with a second-surface mirror.) Explain the reasons for these design differences.

Some reflecting telescope mirrors utilize a rotating tub of mercury to produce a large parabolic surface. If the tub is rotating on its axis with an angular frequency \(\omega,\) show that the focal length of the resulting mirror is: \(f=g / 2 \omega^{2}\).

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?

For a person whose near point is \(115 \mathrm{~cm},\) so that he can read a computer monitor at \(55 \mathrm{~cm},\) what power of read ing glasses should his optician prescribe, keeping the lenseye distance of \(2.0 \mathrm{~cm}\) for his spectacles?

A classmate claims that by using a \(40.0-\mathrm{cm}\) focal length mirror, he can project onto a screen a \(10.0-\mathrm{cm}\) tall bird locat ed 100 . \(\mathrm{m}\) away. He claims that the image will be no less than \(1.00 \mathrm{~cm}\) tall and inverted. Will he make good on his claim?
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