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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 鈮� 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

Kim says that she was less than 10 ft away from the president when she took a picture of him with her \(50-\mathrm{mm}\) focal length camera lens. The picture shows the upper half of the president's body (or \(3.0 \mathrm{ft}\) of his total height). On the negative of the film, this part of his body is $18 \mathrm{mm}$ high. How close was Kim to the president when she took the picture?

An object is placed \(20.0 \mathrm{cm}\) from a converging lens with focal length \(15.0 \mathrm{cm}\) (see the figure, not drawn to scale). A concave mirror with focal length \(10.0 \mathrm{cm}\) is located \(75.0 \mathrm{cm}\) to the right of the lens. (a) Describe the final image- -is it real or virtual? Upright or inverted? (b) What is the location of the final image? (c) What is the total transverse magnification?
Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
A microscope has an eyepicce that gives an angular magnification of 5.00 for a final image at infinity and an objective lens of focal length $15.0 \mathrm{mm}\(. The tube length of the microscope is \)16.0 \mathrm{cm} .$ (a) What is the transverse magnification due to the objective lens alone? (b) What is the angular magnification due to the microscope? (c) How far from the objective should the object be placed?
A slide projector has a lens of focal length \(12 \mathrm{cm} .\) Each slide is \(24 \mathrm{mm}\) by \(36 \mathrm{mm}\) (see the figure with Problem 16). The projector is used in a room where the screen is \(5.0 \mathrm{m}\) from the projector. How large must the screen be?
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