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Chapter 24: Problem 20

A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)

Short Answer

Expert verified
Based on the given information, the image location for part (a) is virtual and cannot be projected onto a screen. For part (b), the screen should be placed approximately 3.33 cm to the right of the diverging lens to display a real image.

Step by step solution

01

Identify known values

We are given the focal lengths and positions of two lenses: a converging lens (F1 = 3 cm) placed 4 cm to the right of an object, and a diverging lens (F2 = -5 cm) placed 17 cm to the right of the converging lens for part (a).
02

Calculate the image distance (di) using the Lensmaker's Equation

For the converging lens, we can use the Lensmaker's Equation, which states: (1/f) = (1/do) + (1/di) where f is the focal length, do is the object distance, and di is the image distance. We can find the image distance for the converging lens: (1/3) = (1/4) + (1/di1) Solving for di1, we get: di1 = 12 cm
03

Calculate the new object distance for the diverging lens

Now, we need to calculate the object distance for the diverging lens. As the diverging lens is placed 17 cm to the right of the converging lens, the object distance for the diverging lens is: do2 = 12 cm - 17 cm = -5 cm As the object distance is negative, it means the converging lens produced a virtual and not a real image. However, we will proceed and see if the diverging lens can produce a real image.
04

Calculate the image distance (di) for the diverging lens

Now, we can use the Lensmaker's Equation for the diverging lens: (1/-5) = (1/-5) + (1/di2) Solving for di2, we get: di2 = -5 cm The final image distance is also negative, meaning that the image is virtual and cannot be projected onto a screen. Therefore, there is no location for the screen in part (a).
05

Calculate the new object distance for the diverging lens (part b)

In part (b), the distance between the two lenses is 10 cm. So, the object distance for the diverging lens is now: do2 = 12 cm - 10 cm = 2 cm
06

Calculate the image distance (di) for the diverging lens (part b)

Now, we need to find the image distance produced by the diverging lens. Again, use the Lensmaker's Equation: (1/-5) = (1/2) + (1/di2) Solving for di2, we get: di2 = 10/3 cm ≈ 3.33 cm As the result is positive, the diverging lens forms a real image. The screen can be placed 3.33 cm to the right of the diverging lens to display the image.

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Most popular questions from this chapter

A dissecting microscope is designed to have a large distance between the object and the objective lens. Suppose the focal length of the objective of a dissecting microscope is \(5.0 \mathrm{cm},\) the focal length of the eyepiece is \(4.0 \mathrm{cm},\) and the distance between the lenses is $32.0 \mathrm{cm} .$ (a) What is the distance between the object and the objective lens? (b) What is the angular magnification?
You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high. what is the image height?
A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?
An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$
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