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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 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.
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?
Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$
A converging lens and a diverging lens, separated by a distance of $30.0 \mathrm{cm},$ are used in combination. The converging lens has a focal length of \(15.0 \mathrm{cm} .\) The diverging lens is of unknown focal length. An object is placed \(20.0 \mathrm{cm}\) in front of the converging lens; the final image is virtual and is formed \(12.0 \mathrm{cm}\) before the diverging lens. What is the focal length of the diverging lens?
A camera uses a 200.0 -mm focal length telephoto lens to take pictures from a distance of infinity to as close as 2.0 \(\mathrm{m}\). What are the minimum and maximum distances from the lens to the film?
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