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Chapter 24: Problem 52

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?

Short Answer

Expert verified
Answer: (a) The distance between the lenses (l) is calculated using the lens formula and the given values of the focal lengths of the objective and eyepiece lenses. (b) The diameter of the image produced by the objective (D_image) is found using the relationship between the distances and the diameters of the image and the moon. (c) The angular magnification (M) is determined by the ratio of the focal lengths of the objective and eyepiece lenses.

Step by step solution

01

Find the distance between the lenses (l)

To find the distance between the lenses (l), we will use the lens formula: $$ \frac{1}{f} = \frac{1}{D_o} + \frac{1}{D_i} $$ where f is the focal length of the objective lens, \(D_o\) is the distance from the objective to the image, \(D_i\) is the distance from the eyepiece to the image. We are given \(f_o = 2.4m\) and \(f_e = 0.16m\). We can define \(l = D_o + D_i\) and we know that the image is formed at the focal length of the eyepiece so \(D_i = f_e\). So the equation becomes: $$ \frac{1}{2.4} = \frac{1}{D_o} + \frac{1}{0.16} $$
02

Calculate the distance from the objective lens to the image (\(D_o\))

To find \(D_o\), we will solve the equation we obtained in step 1: $$ \frac{1}{D_o} = \frac{1}{2.4} - \frac{1}{0.16} $$ $$ D_o = \frac{1}{\frac{1}{2.4} - \frac{1}{0.16}} $$ Calculate the value of \(D_o\).
03

Find the distance between the lenses (l)

Now that we have \(D_o\), we can find the distance between the lenses (l) using the relationship \(l = D_o + D_i\): $$ l = D_o + f_e $$ Calculate the value of l.
04

Find the diameter of the image produced by the objective

To find the diameter of the image produced by the objective, we can use the relationship: $$ \frac{D_o}{f_o} = \frac{D_{image}}{D_{moon}} $$ where \(D_{image}\) is the diameter of the image produced by the objective and \(D_{moon}\) is the diameter of the moon. We know that the diameter of the moon is \(3.474 \times 10^6\mathrm{m}\). Now we can solve for \(D_{image}\): $$ D_{image} = \frac{D_o \times D_{moon}}{f_o} $$ Calculate the value of \(D_{image}\).
05

Find the angular magnification

The angular magnification (M) of the telescope is given by the ratio of the focal lengths of the objective and eyepiece lenses: $$ M = \frac{f_o}{f_e} $$ Calculate the value of M.

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Most popular questions from this chapter

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?

A converging lens with a focal length of \(15.0 \mathrm{cm}\) and a diverging lens are placed \(25.0 \mathrm{cm}\) apart, with the converging lens on the left. A 2.00 -cm-high object is placed \(22.0 \mathrm{cm}\) to the left of the converging lens. The final image is \(34.0 \mathrm{cm}\) to the left of the converging lens. (a) What is the focal length of the diverging lens? (b) What is the height of the final image? (c) Is the final image upright or inverted?
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?
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?
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