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Chapter 39: Problem 30

An astronomical telescope with an objective lens of focal length \(+80 \mathrm{~cm}\) is focused on the moon. By how much must the eyepiece be moved to focus the telescope on an object 40 meters distant?

Short Answer

Expert verified
Move the eyepiece by 1.63 cm.

Step by step solution

01

Understand the Problem

We need to determine the shift in the eyepiece position of a telescope initially focused on the moon when the focus is changed to an object 40 meters away. The focal length of the objective lens is given as \( f_o = +80 \, \mathrm{cm} \). The initial object (the moon) is considered at infinity.
02

Calculate Initial Image Position

When the telescope is focused on the moon, the object is considered at infinity. Therefore, the image is formed at the focal point of the objective lens. The initial image distance is equal to the focal length, which is \( v_o = f_o = 80 \, \mathrm{cm} \).
03

Calculate New Object Distance

Next, convert the new object distance from meters to centimeters: \( s_n = 40 \, \text{meters} = 4000 \, \mathrm{cm} \). Use this value for the new object distance in further calculations.
04

Use Lens Formula for New Image Position

Use the lens formula to find the new image distance \( v_n \) for the objective lens:\[ \frac{1}{v_n} = \frac{1}{f_o} - \frac{1}{s_n} \]Substitute the values:\[ \frac{1}{v_n} = \frac{1}{80} - \frac{1}{4000} \]Calculate \( v_n \).
05

Simplify and Solve for New Image Position

Simplify the equation:First, calculate common denominator:\[ \frac{1}{80} = \frac{50}{4000} \text{ (for } 1/4000) \]Now replace in the equation:\[ \frac{1}{v_n} = \frac{50}{4000} - \frac{1}{4000} \]\[ \frac{1}{v_n} = \frac{49}{4000} \]Therefore, \( v_n = \frac{4000}{49} \approx 81.63 \, \mathrm{cm} \).
06

Determine the Shift in Eyepiece Position

Find the change in position of the image, which corresponds to the shift needed in the eyepiece:\[ \Delta v = v_n - v_o = 81.63 - 80 = 1.63 \, \mathrm{cm} \]Hence, the eyepiece must be moved by approximately 1.63 cm.

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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 fundamental property of a lens. It is the distance between the lens and the image formed when the object is situated at infinity. For the telescope's objective lens in this exercise, the focal length is given as +80 cm. This means when an object is placed far away, like the moon, the image will be 80 cm away from the lens.
Focal length plays a critical role in determining how lenses focus light. In general:
  • A long focal length means the lens brings parallel rays of light (like from distant objects) to focus the image farther from the lens.
  • A short focal length bends light more sharply, creating an image closer to the lens.
Understanding focal length helps predict where an image will form and aids in precisely adjusting the telescope.
Lens Formula
The lens formula relates the object distance, image distance, and the focal length of a lens. It is given by:\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]where:
  • \(f\) is the focal length,
  • \(v\) is the image distance, and
  • \(u\) is the object distance.
In our exercise, the telescope is shifted from viewing the moon at infinity to an object 40 meters (4000 cm) away. When focused on the moon, the object distance is considered infinite, simplifying the lens formula to \(v = f\).
For new object distances, the formula allows us to calculate the new image position. By rearranging it, one can find how far inside the lens system the light focuses, providing insights necessary for adjusting the eyepiece.
Image Distance
Image distance is where an image is located concerning the lens. This concept is crucial for focusing telescopes or cameras accurately. Initially, when aiming at a distant object (like the moon), the image forms exactly at the focal length since the object is considered to be infinitely far away.
For this telescope, when focusing on a nearby object 40 meters away, the image distance was calculated using the lens formula.
Previously, the image was formed at 80 cm. With the new object distance, the image shifts slightly, determined to be approximately 81.63 cm away from the objective lens. Understanding image distance helps decide how the eyepiece is adjusted in relation to the lens.
Eyepiece Adjustment
Eyepiece adjustment is essential for achieving clear focus when switching between distant and near objects. Once the new image distance is calculated, we determine how much to move the eyepiece to catch the focused image. This adjustment ensures a sharp and clear image by aligning the eyepiece to the new position where the focused light converges. In this scenario, when moving from the moon to a terrestrial object 40 meters away, the image distance changed slightly, so the eyepiece must be adjusted by about 1.63 cm. This repositioning ensures that the observer views a sharp image, aligning precisely with the altered focal point. Small shifts in the eyepiece are crucial for clear images in varied viewing conditions.

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Most popular questions from this chapter

A nearsighted man cannot see objects clearly that are beyond 50 \(\mathrm{cm}\) from his eye. Determine the focal length and power of the glasses that will enable him to see distant objects clearly.

A farsighted person named Amy cannot see clearly objects closer to the eye than \(75 \mathrm{~cm}\). Determine the power of the spectacle lenses which will enable her to read type at a distance of \(25 \mathrm{~cm}\). The image, which must be right-side-up, must be on the same side of the lens as the type (hence, the image is virtual and \(s_{i}=75 \mathrm{~cm}\) ), and farther from the lens than the type (hence, converging on positive lenses are prescribed). Keep in mind that for virtual images formed by a convex lens \(\left|s_{i}\right|>s_{0}\). We have $$ \frac{1}{f}=\frac{1}{25}-\frac{1}{75} \quad \text { or } \quad f=+37.5 \mathrm{~cm} $$ and $$ \text { Power }=\frac{1}{0.375 \mathrm{~m}}=2.7 \text { diopters } $$

Two positive lenses, having focal lengths of \(+2.0 \mathrm{~cm}\) and \(+5.0 \mathrm{~cm}\), are \(14 \mathrm{~cm}\) apart as shown. An object \(A B\) is placed \(3.0\) \(\mathrm{cm}\) in front of the \(+2.0\) lens. Determine the position and magnification of the final image A"B" formed by this combination of lenses. To locate image \(A^{\prime} B^{\prime}\) formed by the \(+2.0\) lens alone: $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{2.0}-\frac{1}{3.0}=\frac{1}{6.0} \quad \text { or } \quad s_{i}=6.0 \mathrm{~cm} $$ The image \(A^{\prime} B^{\prime}\) is real, inverted, and \(6.0 \mathrm{~cm}\) beyond the \(+2.0\) lens. To locate the final image \(A^{\prime \prime} B^{\prime \prime}\) : The image \(A^{\prime} B^{\prime}\) is \((14-6.0) \mathrm{cm}=\) \(8.0 \mathrm{~cm}\) in front of the \(+5.0\) lens and is taken as a real object for the \(+5.0\) lens. $$ \frac{1}{s_{i}}=\frac{1}{5.0}-\frac{1}{8.0} \quad \text { or } \quad s_{i}=13.3 \mathrm{~cm} $$ \(A^{\prime \prime} B^{\prime \prime}\) is real, erect, and \(13 \mathrm{~cm}\) from the \(+5\) lens. Then, $$ M_{T}=\frac{\overline{A^{\prime \prime} B^{\prime \prime}}}{\overline{A B}}=\frac{\overline{A^{\prime} B^{\prime}}}{\overline{A B}} \times \frac{\overline{A^{\prime \prime} B^{\prime \prime}}}{\overline{A^{\prime} B^{\prime}}}=\frac{6.0}{3.0} \times \frac{13.3}{8.0}=3.3 $$ Note that the magnification produced by a combination of lenses is the product of the individual magnifications.

A microscope has two interchangeable objective lenses (3.0 mm and \(7.0 \mathrm{~mm}\) ) and two interchangeable eyepieces \((3.0 \mathrm{~cm}\) and \(5.0\) \(\mathrm{cm}\) ). What magnifications can be obtained with the microscope if it is adjusted so that the image formed by the objective is \(17 \mathrm{~cm}\) from that lens? Because \(s_{i o}=17 \mathrm{~cm}\) the magnification formula for a microscope, with \(d_{n}=25 \mathrm{~cm}\), gives the following possibilities for \(M_{A}\) : For \(f_{E}=3 \mathrm{~cm}, f_{o}=0.3 \mathrm{~cm}:\) For \(f_{E}=3 \mathrm{~cm}, f_{o}=0.7 \mathrm{~cm}:\) For \(f_{g}=5 \mathrm{~cm}, f_{\theta}=0.3 \mathrm{~cm}:\) For \(f_{E}=5 \mathrm{~cm}, f_{o}=0.7 \mathrm{~cm}:\) \(M_{A}=(9.33)(55,6)=518=5.2 \times 10^{2}\) \(M_{A}=(9.33)(23.2)=216=2.2 \times 10^{2}\) \(M_{A}=(5)(55.6)=278=2.8 \times 10^{2}\) \(M_{A}=(5)(23.2)=116=1.2 \times 10^{2}\)

Compute the magnifying power of a telescope, having objective and eyepiece lenses of focal lengths \(+60\) and \(+3.0 \mathrm{~cm}\), respectively, when it is focused for parallel rays. The image is inverted.
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