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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

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?
A refracting telescope is used to view the Moon. The focal lengths of the objective and eyepiece are \(+2.40 \mathrm{m}\) and \(+16.0 \mathrm{cm},\) respectively. (a) What should be the distance between the lenses? (b) What is the diameter of the image produced by the objective? (c) What is the angular magnification?
A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
Suppose the distance from the lens system of the eye (cornea + lens) to the retina is \(18 \mathrm{mm}\). (a) What must the power of the lens be when looking at distant objects? (b) What must the power of the lens be when looking at an object \(20.0 \mathrm{cm}\) from the eye? (c) Suppose that the eye is farsighted; the person cannot see clearly objects that are closer than $1.0 \mathrm{m}$. Find the power of the contact lens you would prescribe so that objects as close as \(20.0 \mathrm{cm}\) can be seen clearly.
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?
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