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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

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 statue is \(6.6 \mathrm{m}\) from the opening of a pinhole camera, and the screen is \(2.8 \mathrm{m}\) from the pinhole. (a) Is the image erect or inverted? (b) What is the magnification of the image? (c) To get a brighter image, we enlarge the pinhole to let more light through, but then the image looks blurry. Why? (d) To admit more light and still have a sharp image, we replace the pinhole with a lens. Should it be a converging or diverging lens? Why? (e) What should the focal length of the lens be?
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 microscope has an eyepiece of focal length \(2.00 \mathrm{cm}\) and an objective of focal length \(3.00 \mathrm{cm} .\) The eyepiece produces a virtual image at the viewer's near point \((25.0 \mathrm{cm}\) from the eye). (a) How far from the eyepiece is the image formed by the objective? (b) If the lenses are \(20.0 \mathrm{cm}\) apart, what is the distance from the objective lens to the object being viewed? (c) What is the angular magnification?
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