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Chapter 33: Problem 63

(II) An astronomical telescope has an objective with focal length \(75 \mathrm{~cm}\) and a \(+35 \mathrm{D}\) eyepiece. What is the total magnification?

Short Answer

Expert verified
The total magnification is approximately -25.86, indicating an inverted image.

Step by step solution

01

Understanding the Focal Lengths

First, we need to understand the information given. The objective lens has a focal length of \(75\, \text{cm}\), which is \(0.75\, \text{m}\). The eyepiece is given in diopters \((\text{D})\), and the focal length \(f\) in meters can be found using \(f = \frac{1}{\text{D}}\). So, for a \(+35\; \text{D}\) eyepiece, the focal length \(f_e = \frac{1}{35}\, \text{m}\).
02

Calculating Focal Length of the Eyepiece

Calculate the focal length of the eyepiece in meters: \[ f_e = \frac{1}{35} = 0.02857\ldots \approx 0.029 \text{ m}. \] This is the focal length of the eyepiece.
03

Using the Magnification Formula

The magnification \(M\) of an astronomical telescope is given by the formula \(M = -\frac{f_o}{f_e}\), where \(f_o\) is the focal length of the objective, and \(f_e\) is the focal length of the eyepiece.
04

Substituting Values into the Formula

Substitute the known values into the magnification formula: \[ M = -\frac{0.75}{0.029}. \] Proceed with the division.
05

Calculating the Magnification

Perform the division: \[ M = -\frac{0.75}{0.029} \approx -25.86. \] The negative sign indicates an inverted image, which is normal for telescopes.

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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 of a lens is the distance from the lens to the point where it focuses parallel rays of light. This point is known as the focal point. The focal length is essential in understanding how lenses work in devices like telescopes and cameras.
For astronomical telescopes, focal length plays a vital role in determining the magnification power. Specifically, the longer the focal length of the objective lens, the greater the magnification potential of the telescope.
In mathematical terms, the focal length (\( f \)) is typically measured in meters (m) or centimeters (cm). In our exercise, the objective lens of the telescope has a focal length of 75 cm, which converts to 0.75 meters. This consistent unit conversion allows for accurate calculations in optics equations.
Objective Lens
The objective lens is the primary lens in an optical device like a telescope. It is the lens that first collects light and begins the process of image formation. The objective lens determines much of the telescope's power due to its larger size compared to the eyepiece.
In a telescope, the objective lens has a long focal length, which helps gather more light and produce a brighter and clearer image. A longer focal length also contributes to a higher level of detail in the observed celestial objects.
For our telescope example, the objective lens has a focal length of 75 cm. This focal length interacts with the eyepiece lens to determine the overall magnification of the telescope. Thus, a precise understanding of the objective focal length is crucial for calculating and enhancing the telescope's performance.
Eyepiece
The eyepiece is another crucial component of an astronomical telescope. It is where the user places their eye to view the image created by the objective lens. Unlike the objective lens, the eyepiece has a much shorter focal length, which increases the magnifying power of the telescope.
The focal length of an eyepiece is often expressed in diopters, especially for lenses with short focal lengths. A diopter (\( D \)) is the unit of measurement for the optical power of a lens and is calculated as the inverse of the focal length in meters (\( f = \frac{1}{D} \)).
In our scenario, the eyepiece has a power of +35 D, resulting in a focal length of approximately 0.029 meters. This short focal length, combined with the longer focal length of the objective lens, results in a magnified image for the observer. Thus, the eyepiece is essential for refining and focusing the image provided by the objective lens, significantly enhancing the user's viewing experience.

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Most popular questions from this chapter

(III) In the "magnification" method, the focal length \(f\) of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances \(d_{i}\) and their associated magnifications \(m\) (minus sign indicates that image is inverted). The data taken in such an experiment are given here:\(\begin{array}{rrrrrr}{d_{\mathrm{i}}(\mathrm{cm})} & {20} & {25} & {30} & {35} & {40} \\ {m} & {-0.43} & {-0.79} & {-1.14} & {-1.50} & {-1.89}\end{array}\) (a) Show analytically that a graph of \(m\) vs. \(d_{\text { i should }}\) produce a straight line. What are the theoretically expected values for the slope and the \(y\) -intercept of this line? [Hint: \(d_{\mathrm{o}}\) is not constant.] \((b)\) Using the data above, graph \(m\) vs. \(d_{\mathrm{i}}\) and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the \(y\) -intercept of your plot have the expected value? (c) In performing such an experiment, one has the practical problem of locating the exact center of the lens since \(d_{\mathrm{i}}\) must be measured from this point. Imagine, instead, that one measures the image distance \(d\) from the back surface of the lens, which is a distance \(\ell\) from the lens's center. Then, \(d_{i}=d_{1}^{\prime}+\ell .\) Show that, when implementing the magnification method in this fashion, a plot of \(m\) vs.di will still result in a straight line. How can \(f\) be determined from this straight line?

A woman can see clearly with her right eye only when objects are between \(45 \mathrm{~cm}\) and \(155 \mathrm{~cm}\) away. Prescription bifocals should have what powers so that she can see distant objects clearly (upper part) and be able to read a book \(25 \mathrm{~cm}\) away (lower part) with her right eye? Assume that the glasses will be \(2.0 \mathrm{~cm}\) from the eye.

(II) A person struggles to read by holding a book at arm's length, a distance of \(55 \mathrm{~cm}\) away. What power of reading glasses should be prescribed for her, assuming they will be placed \(2.0 \mathrm{~cm}\) from the eye and she wants to read at the "normal" near point of \(25 \mathrm{~cm} ?\)

(II) A converging lens has focal length \(f\). When an object is placed a distance \(d_{0}>f\) from this lens, a real image with magnification \(m\) is formed. \((a)\) Show that \(m=f /\left(f-d_{0}\right)\). (b) Sketch \(m\) vs, \(d_{\mathrm{o}}\) over the range \(f

(II) Two 25.0 -cm-focal-length converging lenses are placed \(16.5 \mathrm{~cm}\) apart. An object is placed \(35.0 \mathrm{~cm}\) in front of one lens. Where will the final image formed by the second lens be located? What is the total magnification?
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