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Find an equation for the length \(L\) of a refracting telescope in terms of the focal length of the objective \(f_{0}\) and the magnification \(m .\) (b) \(A\) knob adjusts the eyepiece forward and backward. Suppose the telescope is in focus with an eyepiece giving a magnification of \(50.0 .\) By what distance must the eyepiece be adjusted when the eyepiece is replaced, with a resulting magnification of \(1.00 \times 10^{22}\) Must the eyepiece be adjusted backward or forward? Assume the objective lens has a focal length of \(2.00 \mathrm{~m}\).

Short Answer

Expert verified
The eyepiece must be adjusted backward by a distance of approximately 0.04m.

Step by step solution

01

Identify Known Values

Here we are given: The focal length of the objective lens, \(f_{0} = 2.00m\); The first magnification, \(m_{1} = 50\);The second magnification, \(m_{2} = 1.00 \times 10^{22}\).
02

Formulate the Telescope Magnification Formula

The magnification \(m\) of a telescope is given by the equation \(m = f_{0}/f_{e}\), where \(f_{e}\) is the focal length of the eyepiece.
03

Calculate the Initial Position of the Eyepiece

First, we calculate the original focal length of the eyepiece using the first magnification. Rearranging the formula gives \(f_{e1} = f_{0}/m_{1}\). Substituting the known values results in \(f_{e1} = 2.00m/50 = 0.04m\). This means the eyepiece was originally 0.04m away from the focal point of the objective lens.
04

Calculate the New Position of the Eyepiece

Next, we calculate the new focal length of the eyepiece using the second magnification. Again, rearranging the formula to \(f_{e2} = f_{0}/m_{2}\), we find that \(f_{e2} = 2.00m/1.00 \times 10^{22} = 2.00 \times 10^{-22}m\). This is the new position of the eyepiece from the focal point of the objective lens.
05

Determine the Adjustment Needed

Finally, we compare the two eyepiece positions, \(f_{e1}\) and \(f_{e2}\), to find the distance the eyepiece needs to be adjusted. The adjustment is \(f_{e2} - f_{e1} = 2.00 \times 10^{-22}m - 0.04m = - 0.04m + 2.00 \times 10^{-22}m = -0.04m\). Because this number is negative, the eyepiece must be adjusted backwards towards the objective lens.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Focal Length
The focal length is a crucial concept in understanding how telescopes function. It is the distance between the lens and the point where it brings light rays into focus. For a telescope, there are generally two focal lengths to be concerned with: the focal length of the objective lens and the focal length of the eyepiece lens.
The objective lens's focal length, often denoted as \( f_0 \), is the primary factor that determines the physical size of the telescope. A larger \( f_0 \) means a longer telescope and usually allows more detailed images. In this particular exercise, we're given a focal length for the objective lens of \( f_0 = 2.00 \text{ m} \).
On the other hand, the focal length of the eyepiece, \( f_e \), directly influences the magnification power of the telescope. Using the relationship \( m = \frac{f_0}{f_e} \), where \( m \) is the magnification, we can see how \( f_e \) adjusts with different magnifications and how essential it is to achieve the desired viewing clarity.
Magnification
Magnification is the process of enlarging the appearance of an object, particularly significant in telescopic observations. The magnification of a refracting telescope is determined by the ratio of the focal lengths of the objective lens and the eyepiece. It is expressed by the equation \( m = \frac{f_0}{f_e} \).
This equation shows that a longer focal length of the objective lens \( f_0 \) or a shorter focal length of the eyepiece \( f_e \) will result in greater magnification. For instance, in our exercise, the initial magnification was \( m_1 = 50 \), implying that the objects observed appear 50 times larger than their actual size.
When a change in the magnification is needed, as in altering \( m_1 = 50 \) to \( m_2 = 1.00 \times 10^{22} \), it means changing either \( f_0 \) or \( f_e \). Typically \( f_0 \) is fixed, so adjustments are made to \( f_e \), involving the eyepiece replacement or adjustment to achieve the new desired magnification.
Eyepiece Adjustment
The eyepiece adjustment is a process by which the position of the eyepiece in the telescope is changed to achieve the correct focal point for clear viewing, especially when changing magnifications. This adjustment is necessary because each eyepiece has a different focal length, \( f_e \), which impacts how the telescope should be set up.
Eyepiece adjustment is crucial when replacing the eyepiece or changing magnifications. In our example, we initially had \( f_{e1} \) calculated for a magnification of 50, resulting in the eyepiece being positioned 0.04 m from the focal point. Later, to achieve an extraordinarily high magnification of \( 1.00 \times 10^{22} \), we figured out \( f_{e2} \) would be incredibly small at \( 2.00 \times 10^{-22} \) m.
Therefore, the eyepiece should be adjusted from the initial position of 0.04 m back nearly to the focal point of the objective lens. The calculated move is backward, showing a rare scenario where the adjustment is so small due to an extremely high magnification requirement.

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Most popular questions from this chapter

A patient can't see objects closer than \(40.0 \mathrm{~cm}\) and wishes to clearly see objects that are \(20.0 \mathrm{~cm}\) from his eye. (a) Is the patient nearsighted or farsighted? (b) If the eye-lens distance is \(2.00 \mathrm{~cm}\), what is the minimum object distance \(p\) from the lens? (c) What image position with respect to the lens will allow the patient to see the object? (d) Is the image real or virtual? Is the image distance \(q\) positive or negative? (e) Calculate the required focal length. (f) Find the power of the lens in diopters. (g) If a contact lens is to be prescribed instead, find \(p, q\), and \(f\), and the power of the lens.

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Galileo devised a simple terrestrial telescope that produces an upright image. It consists of a converging objective lens and a diverging eyepiece at opposite ends of the telescope tube. For distant objects, the tube length is the objective focal length less the absolute value of the eyepiece focal length. (a) Does the user of the telescope see a real or virtual image? (b) Where is the final image? (c) If a telescope is to be constructed with a tube of length \(10.0 \mathrm{~cm}\) and a magnification of \(3.00\), what are the focal lengths of the objective and eyepiece?

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