Problem 103 - Physics for Scientists and Engineers with Modern Physics

Question

The objective lens and the eyepiece of a telescope are spaced \(85 \mathrm{~cm}\) apart. If the eyepiece is \(+23 \mathrm{D},\) what is the total magnification of the telescope?

Step-by-Step Solution

Answer

The total magnification of the telescope is approximately 18.54.

Step 1: Understand Telescope Magnification

Telescope magnification is calculated using 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. Given that the objective lens and the eyepiece are spaced 85 cm apart, \( f_o + f_e = 85 \) cm.

Step 2: Convert Eyepiece Power to Focal Length

The focal length \( f_e \) for the eyepiece can be calculated from its power using the formula: \( f_e = \frac{1}{P} \) where \( P \) is the power in diopters. Here, \( P = +23 \) D, so \( f_e = \frac{1}{23} \text{ meters} = \frac{100}{23} \text{ cm} \approx 4.35 \text{ cm}. \)

Step 3: Determine the Objective Lens Focal Length

Using the equation from Step 1: \( f_o + f_e = 85 \text{ cm} \), we substitute the value of \( f_e \) to find \( f_o \). Thus, \( f_o = 85 - 4.35 = 80.65 \text{ cm}. \)

Step 4: Calculate the Magnification

Apply the magnification formula: \( M = \frac{f_o}{f_e} = \frac{80.65}{4.35} \approx 18.54. \) Therefore, the total magnification of the telescope is approximately 18.54.

Key Concepts

Objective Lens

The objective lens is a key component of a telescope. It collects light from a distant object and focuses it to create an image. The purpose of the objective lens is critical because it determines the amount of light that enters the telescope, directly influencing the clarity and brightness of the observed image.

Some critical aspects of the objective lens include:

• Size and Diameter: A larger diameter allows more light to enter, which is beneficial for viewing faint objects.
• Focal Length: The distance from the lens to its focus point, where the image is formed. In our exercise, this is represented by \( f_o \).

Understanding the role of the objective lens helps in comprehending how telescopes function and why their configuration impacts the viewing experience.

Eyepiece

The eyepiece, unlike the objective lens, is used for magnifying the image created by the objective lens. It acts much like a magnifying glass and is responsible for the second stage of magnification in a telescope.

Key elements of the eyepiece include:

• Power: Measured in diopters, this indicates the eyepiece's ability to magnify an image. Higher power results in higher magnification.
• Focal Length: Calculated using its power. It is given by the formula \( f_e = \frac{1}{P} \), where \( P \) is the power in diopters. For instance, in this exercise, the eyepiece has a power of +23D, resulting in \( f_e \) of approximately 4.35 cm.

The eyepiece ultimately influences how large the image appears to the observer, making its focal length a crucial factor in designing telescopes.

Focal Length

The focal length in telescopes refers to the distance over which light rays are brought to focus after passing through a lens. It is fundamental in determining how telescopes form images and how these images are magnified.

Important considerations about focal length are:

• Objective Focal Length: Longer focal lengths in objective lenses result in sharper and clearer images. In the exercise, we find the objective lens has a focal length of 80.65 cm.
• Eyepiece Focal Length: A shorter focal length increases magnification. The eyepiece focal length, calculated based on the power of +23D, is approximately 4.35 cm.
• Magnification: The telescope’s magnification, \( M \), relates to the focal lengths of both the objective and eyepiece, using the formula \( M = \frac{f_o}{f_e} \). This shows how changing focal lengths alters a telescope's ability to magnify an image.

Understanding focal length is vital in optimizing telescope design to satisfy specific observational requirements.