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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

A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.
A camera with a 50.0 -mm lens can focus on objects located from $1.5 \mathrm{m}$ to an infinite distance away by adjusting the distance between the lens and the film. When the focus is changed from that for a distant mountain range to that for a flower bed at \(1.5 \mathrm{m},\) how far does the lens move with respect to the film?
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 slide projector, using slides of width \(5.08 \mathrm{cm},\) produces an image that is \(2.00 \mathrm{m}\) wide on a screen \(3.50 \mathrm{m}\) away. What is the focal length of the projector lens?
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