Problem 89
Question
Mars subtends an angle of \(8.0 \times 10^{-5} \mathrm{rad}\) at the unaided eye. An astronomical telescope has an eyepiece with a focal length of \(0.032 \mathrm{~m}\). When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3} \mathrm{rad}\). Find the focal length of the telescope's objective lens.
Step-by-Step Solution
Verified Answer
The focal length of the telescope's objective lens is 1.12 meters.
1Step 1: Understand the Problem Setup
We need to find the focal length of the telescope's objective lens based on the given angle subtended by Mars at the unaided eye and when viewed through the telescope.
2Step 2: Establish the Telescope Magnification Formula
The angular magnification (or magnifying power) of a telescope is given by the formula \( M = \frac{\theta'}{\theta} = \frac{f_o}{f_e} \), where \( \theta' \) is the angle subtended through the telescope, \( \theta \) is the angle subtended to the unaided eye, \( f_o \) is the focal length of the objective lens, and \( f_e \) is the focal length of the eyepiece.
3Step 3: Substitute Known Values into the Formula
We are given: \( \theta = 8.0 \times 10^{-5} \, \text{rad} \), \( \theta' = 2.8 \times 10^{-3} \, \text{rad} \), and \( f_e = 0.032 \, \text{m} \). Substituting these into the formula yields: \[ M = \frac{2.8 \times 10^{-3}}{8.0 \times 10^{-5}} = \frac{f_o}{0.032} \]
4Step 4: Calculate the Magnification
Calculate the magnifying power \( M \):\[ M = \frac{2.8 \times 10^{-3}}{8.0 \times 10^{-5}} = 35 \]
5Step 5: Solve for the Objective Lens Focal Length
With the calculated magnification, substitute back into the equation \( M = \frac{f_o}{f_e} \) to solve for \( f_o \):\[ 35 = \frac{f_o}{0.032} \]Solve for \( f_o \):\[ f_o = 35 \times 0.032 = 1.12 \, \text{m} \]
6Step 6: Conclusion
The focal length of the telescope's objective lens is 1.12 meters. This means the telescope effectively enhances the angular size of Mars from an unaided view to what is seen as enhanced through the telescope by this objective focal length.
Key Concepts
Focal Length CalculationAngular MagnificationAstronomical Observation
Focal Length Calculation
Calculating the focal length of a telescope's objective lens is essential to understanding how telescopes work. The focal length determines how strongly the lens bends or focuses light.
In our exercise, we need to find the focal length of the objective lens using given angles and the focal length of the eyepiece.
In our exercise, we need to find the focal length of the objective lens using given angles and the focal length of the eyepiece.
- The sky object, like Mars, subtends an angle of \(8.0 \times 10^{-5}\) radians to the unaided eye.
- Using the telescope, the angle becomes \(2.8 \times 10^{-3}\) radians.
- The eyepiece focal length is \(0.032\) meters.
Angular Magnification
Angular magnification, or magnifying power, is a measure of how much larger a telescope can make objects appear. It's crucial in choosing or designing a telescope that fits specific observational needs.The formula for angular magnification is:\[ M = \frac{\theta'}{\theta} = \frac{f_o}{f_e} \]
- \( \theta' \) is the angle seen through the telescope.
- \( \theta \) is the angle to the unaided eye.
- \( f_o \) is the focal length of the objective lens.
- \( f_e \) is the focal length of the eyepiece.
Astronomical Observation
Astronomical observations allow us to explore and study distant celestial bodies like Mars and other planets. Telescopes play a vital role in this exploration by magnifying these distant objects, making them visible or more detailed to the human eye. The focal lengths of the lenses and the angular magnification power are key factors in effective astronomical observations.
By using a telescope:
- The light from celestial objects can be collected and focused efficiently.
- Details not visible to the unaided eye, such as surface features of planets, can be examined.
- Changes or movements in these celestial bodies can be monitored.
Other exercises in this chapter
Problem 87
In a compound microscope, the focal length of the objective is \(3.50 \mathrm{~cm}\) and that of the eyepiece is \(6.50 \mathrm{~cm} .\) The distance between th
View solution Problem 88
In a compound microscope, the objective has a focal length of \(0.60 \mathrm{~cm},\) while the eyepiece has a focal length of \(2.0 \mathrm{~cm} .\) The separat
View solution Problem 89
Mars subtends an angle of \(8.0 \times 10^{-5}\) rad at the unaided eye. An astronomical telescope has an eyepiece with a focal length of \(0.032 \mathrm{~m}\).
View solution Problem 90
An astronomical telescope for hobbyists has an angular magnification of \(-155 .\) The eyepiece has a focal length of \(5.00 \mathrm{~mm}\). (a) Determine the f
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