Problem 92
Question
An astronomical telescope has an angular magnification of \(-132 .\) Its objective has a refractive power of 1.50 diopters. What is the refractive power of its eyepiece?
Step-by-Step Solution
Verified Answer
The refractive power of the eyepiece is 198 diopters.
1Step 1: Understand Angular Magnification Formula
The angular magnification of a telescope is given by the formula \( M = - \frac{F_o}{F_e} \), where \( M \) is the angular magnification, \( F_o \) is the focal length of the objective lens, and \( F_e \) is the focal length of the eyepiece. The negative sign indicates that the image is inverted.
2Step 2: Relate Refractive Power to Focal Length
Refractive power (in diopters) is the reciprocal of the focal length in meters, i.e., \( F = \frac{1}{f} \). Therefore, the focal length can be found as \( f = \frac{1}{F} \). The objective's refractive power is \(1.50\) diopters, giving it a focal length \( f_o = \frac{1}{1.50} \) meters.
3Step 3: Calculate Focal Length of Eyepiece
Using the angular magnification formula, substitute \( M = -132 \) and \( F_o = \frac{1}{1.50} \). Solve for the focal length of the eyepiece \( f_e \): \[ -132 = - \frac{\frac{1}{1.50}}{f_e} \] Rearranging gives: \[ 132 \cdot f_e = \frac{1}{1.50} \]\[ f_e = \frac{1}{132 \cdot 1.50} \]
4Step 4: Calculate Refractive Power of Eyepiece
Find the refractive power of the eyepiece using \( F_e = \frac{1}{f_e} \). Substitute the value of \( f_e \) from the previous step:\[ F_e = \frac{1}{\frac{1}{132 \cdot 1.50}} = 132 \cdot 1.50 \] \( F_e = 198 \, \text{diopters} \).
Key Concepts
Angular MagnificationRefractive PowerFocal Length
Angular Magnification
Angular magnification is a crucial concept when discussing optical instruments like telescopes. In the case of an astronomical telescope, the angular magnification (\(M\)) tells us how much larger (or smaller) an object appears when viewed through the telescope compared to the naked eye.
- The formula for angular magnification in telescopes is \(M = -\frac{F_o}{F_e}\).
- In this formula, \(F_o\) represents the focal length of the objective lens, and \(F_e\) represents the focal length of the eyepiece.
- The negative sign indicates that the image is inverted, which is typical for telescopes.
Refractive Power
Refractive power is another foundational concept in optics, especially relevant when dealing with lenses. It describes how strongly a lens converges (focuses) or diverges light. The refractive power (\(F\)) is measured in diopters and is determined by the reciprocal of the focal length in meters:
- \(F = \frac{1}{f}\), where \(f\) is the focal length in meters.
Focal Length
The focal length of a lens is a distance from the lens where parallel rays of light converge to a point. It is a critical factor in determining both the magnification and refractive power of lenses used in telescopes. The focal length influences:
- The field of view and the amount of detail that can be seen.
- The overall length of the telescope system, as it adds the objective and eyepiece focal lengths.
- The clarity of the image — longer focal lengths generally produce clearer images.
Other exercises in this chapter
Problem 90
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