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Chapter 24: Problem 1

An object is placed \(12.0 \mathrm{cm}\) in front of a lens of focal length $5.0 \mathrm{cm} .\( Another lens of focal length \)4.0 \mathrm{cm}\( is placed \)2.0 \mathrm{cm}$ past the first lens. (a) Where is the final image? Is it real or virtual? (b) What is the overall magnification? (interactive: virtual optics lab).

Short Answer

Expert verified
(a) The final image is a virtual image, located 44 cm from the second lens on the object's side. (b) To find the overall magnification, use the magnifications calculated in steps 3 and 6: \[ M = M_1 \times M_2 \] Substitute the values and calculate the overall magnification.

Step by step solution

01

Use the Thin Lens Equation for the First Lens

First, we need to find the image position and magnification produced by the first lens of focal length \(5.0 \mathrm{cm}\). We will use the thin lens equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. Plugging in our values: \[ \frac{1}{5} = \frac{1}{12} + \frac{1}{d_i} \]
02

Calculate the Image Distance for the First Lens

Solve the equation from step 1 for \(d_i\): \[ d_i = \frac{1}{\frac{1}{5} - \frac{1}{12}} \] \[ d_i = 7.5 \mathrm{cm} \] The image distance for the first lens is \(7.5\,\text{cm}\).
03

Calculate the Magnification for the First Lens

Now, we need to find the magnification (\(M_1\)) of this first image using the following equation: \[ M_1 = -\frac{d_i}{d_o} \] \[ M_1 = -\frac{7.5}{12} \]
04

Use the Second Lens' Thin Lens Equation

The image produced by the first lens will serve as the object for the second lens, which is \(2.0\,\text{cm}\) apart from the first lens. Thus, the new object distance for the second lens is \(7.5 - 2.0 = 5.5 \mathrm{cm}\). Now, we use the thin lens equation for the second lens with a focal length of \(4.0\,\text{cm}\). \[ \frac{1}{4} = \frac{1}{5.5} + \frac{1}{d_{i2}} \]
05

Calculate the Image Distance for the Second Lens

Solve the equation from step 4 for \(d_{i2}\): \[ d_{i2} = \frac{1}{\frac{1}{4} - \frac{1}{5.5}} \] \[ d_{i2} = -44 \mathrm{cm} \] The image distance for the second lens is \(-44\,\text{cm}\). Since it is negative, it means the image is on the same side as the object, making it a virtual image.
06

Calculate the Magnification for the Second Lens

Now, we need to find the magnification (\(M_2\)) produced by the second lens using the following equation: \[ M_2 = -\frac{d_{i2}}{d_{o2}} \] \[ M_2 = -\frac{-44}{5.5} \] Now we have the magnification for both lenses, and the final image is a virtual image.
07

Calculate the Overall Magnification

The overall magnification is the product of the magnifications of both lenses: \[ M = M_1 \times M_2 \] We can now provide the answers for both parts: (a) The final image is \(44\,\text{cm}\) from the second lens on the same side as the object, making it a virtual image. (b) The overall magnification is the product of the magnifications of both lenses, found in steps 3 and 6.

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Most popular questions from this chapter

You have two lenses of focal length \(25.0 \mathrm{cm}\) (lens 1 ) and $5.0 \mathrm{cm}$ (lens 2 ). (a) To build an astronomical telescope that gives an angular magnification of \(5.0,\) how should you use the lenses (which for objective and which for eyepiece)? Explain. (b) How far apart should they be?
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.
Repeat Problem \(40(\mathrm{c})\) using a different eyepiece that gives an angular magnification of 5.00 for a final image at the viewer's near point \((25.0 \mathrm{cm})\) instead of at infinity.
You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length $3.70 \mathrm{cm}\( located \)6.00 \mathrm{cm}$ to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) 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?
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