Problem 94 - Physics

Question

A refracting telescope has an angular magnification of \(-83.00 .\) The length of the barrel is \(1.500 \mathrm{~m}\). What are the focal lengths of (a) the objective and (b) the eyepiece?

Step-by-Step Solution

Answer

Objective focal length: 1482.38 mm; Eyepiece focal length: 17.86 mm.

Step 1: Understand Angular Magnification

The angular magnification \( M \) of a refracting telescope is given by the formula \( M = -\frac{f_o}{f_e} \), where \( f_o \) is the focal length of the objective lens, and \( f_e \) is the focal length of the eyepiece. In this problem, \( M = -83.0 \). This will be used to find the focal lengths.

Step 2: Relate Telescope Length to Focal Lengths

The total length of the telescope barrel \( L \) is the sum of the focal lengths of the objective and the eyepiece: \( L = f_o + f_e \). Here, \( L = 1.5 \) m or \( 1500 \) mm, as measurements are usually consistent in units.

Step 3: Set Up Equations

Write the two equations using the formulas from Steps 1 and 2:1. \( M = -\frac{f_o}{f_e} \) becomes \( -83.0 = -\frac{f_o}{f_e} \).2. \( L = f_o + f_e \) becomes \( 1500 = f_o + f_e \).

Step 4: Solve for \(f_o\) in Terms of \(f_e\)

From the magnification equation, express \( f_o \) in terms of \( f_e \):\[ f_o = 83.0 \times f_e \].

Step 5: Substitute into the Length Equation

Substitute the expression for \( f_o \) from Step 4 into the length equation:\[ 1500 = 83.0 \times f_e + f_e \]Simplify to get:\[ 1500 = 84.0 \times f_e \].

Step 6: Solve for \(f_e\)

Solve the equation from Step 5 for \( f_e \):\[ f_e = \frac{1500}{84} \approx 17.86 \text{ mm} \].

Step 7: Calculate \(f_o\) Using \(f_e\)

Using the value of \( f_e \) from Step 6, find \( f_o \):\[ f_o = 83.0 \times 17.86 \approx 1482.38 \text{ mm} \].

Step 8: Conclusion

The focal lengths of the objective and the eyepiece are approximately \( 1482.38 \text{ mm} \) and \( 17.86 \text{ mm} \), respectively.

Key Concepts

Angular Magnification

Angular magnification is a key concept in the functioning of refracting telescopes. It describes how much larger an object appears through the telescope compared to how it looks to the naked eye. The formula for angular magnification (\( M \)) in a refracting telescope is:

• \( M = -\frac{f_o}{f_e} \)
• Where \( f_o \) 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. In the example from the original exercise, the angular magnification is \(-83.0\). This means that the telescope makes things look 83 times larger, albeit upside down.
Understanding this concept is crucial as it determines how effectively a telescope can magnify distant objects. The larger the magnification, the more powerful the telescope in terms of vision.

Focal Length

The concept of focal length plays a significant role in the design and function of lenses in optical devices like telescopes. Focal length is defined as the distance between the lens and the point where light rays converge to form a sharp image.It is a crucial factor in determining the magnifying power and clarity of an image.

• A longer focal length of the objective lens (\( f_o \)) leads to higher magnification and helps to capture more light.
• A shorter focal length in the eyepiece (\( f_e \)) results in a smaller, but brighter, image.

In our refracting telescope exercise, these two focal lengths have been calculated based on a 1.500 m barrel length, using the relationship \( L = f_o + f_e \) where \( L \) is the total length of the telescope.
Calculating precise focal lengths ensures optimal performance of the telescope, allowing viewers to see distant astronomical objects with clarity.

Objective Lens

The objective lens is one of the two critical components in a refracting telescope. Its primary job is to collect light from a distant object and create an image at its focal point. The larger and longer the focal length of the objective lens, the more light can be gathered.

• The size and efficiency of the objective lens determine how much detail can be observed.
• In advanced telescopes, the objective lens is what allows for deep space observation.

In the given exercise, the objective lens has a focal length calculated to be approximately 1482.38 mm. This means it gathers light from afar and focuses it sharply, allowing the eyepiece to magnify it further.
Choosing an appropriate objective lens is crucial for achieving the intended magnification and for viewing faint or faraway celestial objects efficiently.

Eyepiece

The eyepiece is the part of the telescope where viewers look through. While the objective lens forms the initial image, the eyepiece magnifies this image so that it can be seen in greater detail.

• The focal length of the eyepiece is essential for determining angular magnification.
• A shorter focal length in the eyepiece increases the overall magnification power of the telescope.

In the case examined in the exercise, the focal length of the eyepiece was calculated to be approximately 17.86 mm.
This relatively short focal length in combination with the objective lens allows for the impressive magnification of \(-83.0\). The eyepiece's role is vital in fine-tuning the viewer's experience, making sure that the view from the telescope is clear and well-magnified.