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Chapter 24: Problem 64

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?

Short Answer

Expert verified
Question: Calculate the intermediate and final image distances, total magnification, and the height of the final image for a combination of two lenses with object distance \(d_o = 1.8cm\), focal lengths \(f_1= +30.0cm\) and \(f_2 = -15.0cm\), and object height \(h_o = 2.0mm\), considering the distance between the lenses is 21cm.

Step by step solution

01

Determine the intermediate image distance d_1'

We are given the object distance \(d_o = 1.8cm\), and the focal length \(f_1= +30.0cm\). To find the intermediate image distance, we'll use the thin lens equation given by \(\frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_1'}\). By plugging in the given values, we'll solve for \(d_1'\): $$ \frac{1}{30} = \frac{1}{1.8} + \frac{1}{d_1'}.$$
02

Calculate the object distance d_2o for the second lens

The object distance for the second lens is the distance between the intermediate image and the second lens, which can be calculated using the distance between the lenses (21cm) and the image distance of the first lens. \(d_2o = 21 - d_1'\).
03

Determine the final image distance d_2'

Using the focal length of the second lens \(f_2 = -15.0cm\), and the object distance \(d_2o\) calculated in step 2, we can find the final image distance using the thin lens equation: \(\frac{1}{f_2} = \frac{1}{d_2o} + \frac{1}{d_2'}\). Solve for \(d_2'\).
04

Calculate the magnification of the first lens M_1

The magnification of the first lens can be found using the formula \(M_1 = -\frac{d_1'}{d_o}\). Substitute the values of \(d_1'\) and \(d_o\) in this formula to find \(M_1\).
05

Calculate the magnification of the second lens M_2

The magnification of the second lens can be found using the formula \(M_2 = -\frac{d_2'}{d_2o}\). Substitute the values of \(d_2'\) and \(d_2o\) in this formula to find \(M_2\).
06

Calculate the total magnification M

The total magnification is the product of the magnifications of two lenses, i.e., \(M = M_1 \times M_2\). Substitute the magnifications \(M_1\) and \(M_2\) from steps 4 and 5 into the formula to find the total magnification.
07

Calculate the height of the final image h_f

Using the total magnification \(M\) and the height of the object \(h_o = 2.0mm\), we can find the height of the final image with the formula \(h_f = M \times h_o\). Multiply the total magnification by the object height to find the height of the final image.

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Most popular questions from this chapter

Comprehensive Problems Good lenses used in cameras and other optical devices are actually compound lenses, made of five or more lenses put together to minimize distortions, including chromatic aberration. Suppose a converging lens with a focal length of \(4.00 \mathrm{cm}\) is placed right next to a diverging lens with focal length of \(-20.0 \mathrm{cm} .\) An object is placed \(2.50 \mathrm{m}\) to the left of this combination. (a) Where will the image be located? (b) Is the image real or virtual?
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?
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.
The distance from the lens system (cornea + lens) of a particular eye to the retina is \(1.75 \mathrm{cm} .\) What is the focal length of the lens system when the eye produces a clear image of an object \(25.0 \mathrm{cm}\) away?
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?
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