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Chapter 24: Problem 5

An object is located \(16.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens, separated by a distance of \(20.0 \mathrm{cm},\) is a diverging lens of focal length \(-10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify using the lens equations.

Short Answer

Expert verified
Answer: The final image is located 40.0 cm to the right of the converging lens.

Step by step solution

01

Calculate the image location after the converging lens

We have a converging lens with focal length \(f_1 = 12.0 cm\) and an object distance \(d_{01} = 16.0 cm\). We can calculate the image distance \(d_{i1}\) using the lensmaker equation: $$\frac{1}{f_1} = \frac{1}{d_{o1}} + \frac{1}{d_{i1}}$$ Rearranging the equation and solving for \(d_{i1}\), we get: $$\frac{1}{d_{i1}} = \frac{1}{f_1} - \frac{1}{d_{o1}}$$ Plug in the given values: $$\frac{1}{d_{i1}} = \frac{1}{12.0 cm} - \frac{1}{16.0 cm}$$ Now, solve for \(d_{i1}\): $$d_{i1} = \frac{1}{\frac{1}{12.0} - \frac{1}{16.0}} = 48.0 \mathrm{cm}$$
02

Calculate the object distance for the diverging lens

Now that we know the image distance (\(d_{i1}\)) after the converging lens, we can calculate the object distance (\(d_{o2}\)) for the diverging lens. The given distance between the lenses is \(20.0 \mathrm{cm}\). Therefore, the object distance for the diverging lens is: $$d_{o2} = d_{i1} - D = 48.0 \mathrm{cm} - 20.0 \mathrm{cm} = 28.0 \mathrm{cm}$$
03

Calculate the image location after the diverging lens

We have a diverging lens with focal length \(f_2 = -10.0 cm\) and object distance \(d_{o2} = 28.0 cm\). We can calculate the image distance \(d_{i2}\) using the lensmaker equation again: $$\frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}}$$ Rearranging the equation and solving for \(d_{i2}\), we get: $$\frac{1}{d_{i2}} = \frac{1}{f_2} - \frac{1}{d_{o2}}$$ Plug in the given values: $$\frac{1}{d_{i2}} = \frac{1}{-10.0 cm} - \frac{1}{28.0 cm}$$ Now, solve for \(d_{i2}\): $$d_{i2} = \frac{1}{\frac{1}{-10.0} - \frac{1}{28.0}} = 20.0 \mathrm{cm}$$
04

Determine the location of the final image

Since the image after the diverging lens is \(d_{i2} = 20.0 \mathrm{cm}\) from the lens, we can calculate the total image distance from the converging lens as follows: $$d_i = D + d_{i2} = 20.0 \mathrm{cm} + 20.0 \mathrm{cm} = 40.0 \mathrm{cm}$$ This is the location of the final image. It is formed \(40.0 \mathrm{cm}\) to the right of the converging lens.

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Most popular questions from this chapter

A converging lens and a diverging lens, separated by a distance of $30.0 \mathrm{cm},$ are used in combination. The converging lens has a focal length of \(15.0 \mathrm{cm} .\) The diverging lens is of unknown focal length. An object is placed \(20.0 \mathrm{cm}\) in front of the converging lens; the final image is virtual and is formed \(12.0 \mathrm{cm}\) before the diverging lens. What is the focal length of the diverging lens?
(a) If Harry has a near point of \(1.5 \mathrm{m},\) what focal length contact lenses does he require? (b) What is the power of these lenses in diopters?
Show that if two thin lenses are close together \((s,\) the distance between the lenses, is negligibly small), the two lenses can be replaced by a single equivalent lens with focal length \(f_{\mathrm{eq}} .\) Find the value of \(f_{\mathrm{eq}}\) in terms of \(f_{1}\) and \(f_{2}.\)
Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find the intermediate and final image distances relative to the corresponding lenses. (b) What is the total magnification? (c) What is the height of the final image?
Two lenses, separated by a distance of \(21.0 \mathrm{cm},\) are used in combination. The first lens has a focal length of \(+30.0 \mathrm{cm} ;\) the second has a focal length of \(-15.0 \mathrm{cm} .\) An object, $2.0 \mathrm{mm}\( long, is placed \)1.8 \mathrm{cm}$ before the first lens. (a) What are the intermediate and final image distances relative to the corresponding lenses? (b) What is the total magnification? (c) What is the height of the final image?
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