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Chapter 24: Problem 61

Two lenses, of focal lengths \(3.0 \mathrm{cm}\) and \(30.0 \mathrm{cm},\) are used to build a small telescope. (a) Which lens should be the objective? (b) What is the angular magnification? (c) How far apart are the two lenses in the telescope?

Short Answer

Expert verified
In a small telescope with lenses having focal lengths of 3.0 cm and 30.0 cm, the lens with a 30 cm focal length should be the objective lens. The total angular magnification of the telescope is approximately 9.33. The distance between the objective and eyepiece lenses in the telescope is 33 cm.

Step by step solution

01

Determine the objective lens

In a telescope, the objective lens is the one with a larger focal length because it can capture more light and provides a magnified image of the distant object. In this case, the lens with focal length 30 cm has a longer focal length than the 3 cm lens. So, the 30 cm lens should be the objective lens.
02

Find the magnification of the objective lens

To do this, we need to use the lens equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) Here, \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. For the objective lens (30 cm), we can assume that the object distance is at infinity since we are looking at distant objects, so \(d_o \to \infty\). In this case, the lens equation becomes: \( \frac{1}{f} = \frac{1}{d_i} \) Solving for the image distance \(d_i\) for the objective lens: \( d_i = f = 30\,\text{cm} \) Now, the magnification of the objective lens can be calculated as: \( M_1 = -\frac{d_i}{d_o} \) Since \(d_o \to \infty\) , \(M_1 \approx 0\).
03

Find the magnification of the eyepiece lens

The eyepiece lens (3 cm) acts as a simple magnifying glass, with the image produced by the objective lens as its object. The angular magnification of the eyepiece focal length can be calculated as: \(M_2 = 1 + \frac{D}{f_{\mathrm{eyepiece}}}\) Here, \(D\) is the near point distance of the eye, typically assumed to be 25 cm for a relaxed eye. Therefore, \(M_2 = 1 + \frac{25\,\text{cm}}{3\,\text{cm}} = 1 + 8.33 = 9.33\)
04

Calculate the total angular magnification

Now, we can calculate the total angular magnification for the telescope as the product of the magnifications of the objective lens and eyepiece lens. \(M_{\mathrm{total}} = M_1 \cdot M_2 \approx 9.33\) So, the angular magnification of the telescope is approximately 9.33.
05

Determine the distance between the lenses

Lastly, we will find the distance between the two lenses. Since we already know the image distance for the objective lens (30 cm), the distance between the lenses will be equal to the image distance of the objective lens plus the focal length of the eyepiece lens: \( d = d_i + f_{\mathrm{eyepiece}} = 30\,\text{cm} + 3\,\text{cm} = 33\,\text{cm} \) So, the distance between the two lenses in the telescope is 33 cm.

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Most popular questions from this chapter

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?
An object is placed \(12.0 \mathrm{cm}\) in front of a lens of focal length $5.0 \mathrm{cm} .\( Another lens of focal length \)4.0 \mathrm{cm}\( is placed \)2.0 \mathrm{cm}$ past the first lens. (a) Where is the final image? Is it real or virtual? (b) What is the overall magnification? (interactive: virtual optics lab).
A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)
You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high. what is the image height?
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.
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