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(II) An astronomical telescope has its two lenses spaced \(78.0 \mathrm{~cm}\) apart. If the objective lens has a focal length of \(75.5 \mathrm{~cm},\) what is the magnification of this telescope? Assume a relaxed cye.

Short Answer

Expert verified
The magnification of the telescope is 30.2.

Step by step solution

01

Understanding the Problem Statement

We are given that the lenses of an astronomical telescope are spaced 78.0 cm apart, and the focal length of the objective lens is 75.5 cm. We need to find the magnification of the telescope when it is set for a relaxed eye.
02

Identify the Known Values

From the problem, the distance between the lenses is 78.0 cm and the focal length of the objective lens ( f_o ) is 75.5 cm. We need to determine the focal length of the eyepiece ( f_e ).
03

Calculate Focal Length of Eyepiece

Since the total length of the telescope (distance between lenses) is equal to the sum of the focal lengths of the objective and the eyepiece, we have:\[f_o + f_e = 78.0 \]Substituting f_o = 75.5 cm, we find f_e :\[75.5 + f_e = 78.0 \ f_e = 78.0 - 75.5 = 2.5 \]Thus, the focal length of the eyepiece is 2.5 cm.
04

Calculate the Magnification of the Telescope

The magnification ( M ) of an astronomical telescope set for a relaxed eye is given by the formula:\[M = - \frac{f_o}{f_e}\]Substitute f_o = 75.5 cm and f_e = 2.5 cm:\[M = - \frac{75.5}{2.5} = -30.2 \]The negative sign indicates that the image is inverted.
05

Conclusion

The magnification of the telescope is 30.2. When expressed as a modulus, because visual magnification is typically considered in terms of absolute value, the magnification is 30.2 times.

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Key Concepts

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

Objective Lens Focal Length
The focal length of the objective lens in an astronomical telescope is an important factor that determines how much light the telescope can collect and how clearly you can see distant objects. In simple terms, the objective lens is the big lens at the front of the telescope. It collects light from the object you are observing and focuses it to a point inside the tube of the telescope. This focal length, denoted as \(f_o\), is crucial because it helps in forming the initial clear image of the distant object. The longer the focal length of the objective lens, the better it can capture light, which is especially important for observing faint astronomical objects.
In our specific problem, an objective lens has a focal length of 75.5 cm. This means that if parallel rays from a distant star are entering the lens, they will converge at a point 75.5 cm from the lens. Understanding this means grasping how critical the objective lens is for the functionality of the telescope, providing the primary focus needed to observe celestial objects.
Eyepiece Focal Length
The eyepiece lens is another essential part of an astronomical telescope. It serves the role of magnifying the image that is formed by the objective lens. The focal length of the eyepiece, denoted as \(f_e\), plays a pivotal role in determining the size of the image you see.The shorter the focal length of the eyepiece, the more it magnifies the image created by the objective lens. This is key to the overall magnification power of the telescope. The shorter focal length forces your eye to see the image like it is much bigger than it really is.
In our problem, once we've used the objective lens's focal length to capture light and form a clear image, we then use an eyepiece with a focal length of 2.5 cm. Given that the total distance between these lenses is 78.0 cm, this eyepiece's focal length is perfectly calculated to provide a magnification of 30.2 times. Remember, magnification is affected by the relationship between the focal lengths of the objective and eyepiece lenses. Hence, the eyepiece plays a key role in refining the focus for you to see a larger, more detailed image of distant celestial bodies.
Relaxed Eye Observation
A relaxed eye observation setting is vital for comfortable and long-lasting viewing through an astronomical telescope. But what does it mean exactly?
When we talk about relaxed eye observation, we refer to the setup of the telescope where the final image it produces is at infinity. This makes it easier for your eye because it minimizes eye strain, allowing you to focus naturally as if looking into the distance without constant adjustments. For a telescope to be in this mode, the eyepiece is adjusted such that the image produced by the objective lens appears to be at an infinite distance. With relaxed eye observation, viewers can comfortably stargaze for extended periods since their eyes do not have to keep readjusting focus. In the given exercise, the relaxation state ensures that the measured magnification takes into consideration the most comfortable viewing experience, which, in this case, results in a clear, steady magnification of 30.2 times.

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Most popular questions from this chapter

How large is the image of the Sun on film used in a camera with \((a)\) a 28 -mm-focal-length lens, \((b)\) a 50 -mm-focal-length lens, and (c) a 135 -mm- focal-length lens? (d) If the 50 -mm lens is considered normal for this camera, what relative magnification does cach of the other two lenses provide? The Sun has diameter \(1.4 \times 10^{6} \mathrm{~km},\) and it is \(1.5 \times 10^{8} \mathrm{~km}\) away.

(II) A 105 -mm-focal-length lens is used to focus an image on the sensor of a camera. The maximum distance allowed between the lens and the sensor plane is \(132 \mathrm{~mm} .\) (a) How far ahead of the sensor should the lens be if the object to be photographed is \(10.0 \mathrm{~m}\) away? \((b) 3.0 \mathrm{~m}\) away? \((c) 1.0 \mathrm{~m}\) away? (d) What is the closest object this lens could photograph sharply?

The proper functioning of certain optical devices (e.g., optical fibers and spectrometers) requires that the input light be a collection of diverging rays within a cone of halfangle \(\theta\) (Fig. \(33-50\) ). If the light initially exists as a collimated beam (i.e., parallel rays), show that a single lens of focal length \(f\) and diameter \(D\) can be used to create the required input light if \(D / f=2 \tan \theta .\) If \(\theta=3.5^{\circ}\) for a certain spectrometer, what focal length lens should be used if the lens diameter is \(5.0 \mathrm{~cm} ?\)

A converging lens with focal length of \(13.0 \mathrm{~cm}\) is placed in contact with a diverging lens with a focal length of \(-20.0 \mathrm{~cm} .\) What is the focal length of the combination, and is the combination converging or diverging?

(II) An achromatic lens is made of two very thin lenses, placed in contact, that have focal lengths \(f_{1}=-28 \mathrm{~cm}\) and \(f_{2}=+25 \mathrm{~cm} .\) (a) Is the combination converging or diverging? (b) What is the net focal length?

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