Problem 60 - College Physics

Question

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification of the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall and 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Step-by-Step Solution

Answer

(a) Angular magnification is -6.33. (b) Image height is -1.9 cm. (c) Angular size is 0.127 radians.

Step 1: Understand the Problem

A telescope uses lenses to magnify distant objects. In this problem, we have two lenses: an objective lens and an eyepiece lens, with given focal lengths. We need to find the angular magnification, the height of the image formed, and the angular size of the image as viewed through the eyepiece.

Step 2: Angular Magnification Calculation

The angular magnification of a telescope composed of lenses 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 lens. Substituting the given values, \( f_o = 95.0 \text{ cm} \) and \( f_e = 15.0 \text{ cm} \), we find: \[ M = -\frac{95.0}{15.0} = -6.33. \] The negative sign indicates that the image is inverted.

Step 3: Height of Image Formed by Objective

For the objective lens, we use the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance and \( d_i \) is the image distance. Since \( d_o \) is much larger than \( f \), the image is formed near the focal plane of the objective. Therefore, \( d_i \approx f_o = 95.0 \text{ cm} \). The magnification is also given by \( m = -\frac{d_i}{d_o} = -\frac{95.0}{300000} \). The height of the image, \( h_i \), is \( m \times h_o \), with \( h_o = 60.0 \text{ m} = 6000 \text{ cm} \): \[ h_i = -\frac{95.0}{300000} \times 6000 \approx -1.9 \text{ cm}. \]

Step 4: Angular Size of Final Image

The angular size of the image viewed through the eyepiece \( \theta_i \) is given by \( \theta_i = \frac{h_i}{-f_e} \) because the eye is very close to the eyepiece. Substituting the calculated image height and eyepiece focal length: \[ \theta_i = \frac{-1.9}{-15.0} \approx 0.127 \text{ radians}. \]

Key Concepts

Angular Magnification Formula

In the context of telescopes, angular magnification refers to how much larger or smaller the image of an object appears compared to its size when viewed with the naked eye. This magnification is determined using 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 lens. The negative sign indicates that the image is inverted—in other words, it is flipped upside down compared to the original orientation of the object being viewed. For our telescope, substituting the given focal lengths of 95.0 cm for the objective and 15.0 cm for the eyepiece provides an angular magnification of -6.33. This means the image appears 6.33 times larger but upside down compared to viewing it without the telescope.

Lens Formula

The lens formula is essential in optical systems for determining the relationship between the object distance, the image distance, and the focal length of a lens. It is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where \( f \) is the focal length of the lens, \( d_o \) is the distance from the lens to the object, and \( d_i \) is the distance from the lens to the image. This formula is quite versatile and crucial in determining where exactly an image is formed relative to the lens.
In our case, the object, a distant building, is so far from the telescope that we can assume the image is formed at the focal point of the objective lens. This simplifies calculations significantly, allowing us to consider \( d_i \approx f_o \).This approximation is particularly helpful when objects are at a great distance, as in the case of astronomical telescopes or situations where precise image positioning is less critical.

Angular Size

Angular size is a way of describing how large an object appears to an observer, rather than its actual physical size. It measures the angle that an object subtends at the point of observation. The formula for calculating the angular size of an image viewed through the eyepiece is:\[ \theta_i = \frac{h_i}{-f_e} \]where \( \theta_i \) is the angular size, \( h_i \) is the height of the image formed by the objective lens, and \( f_e \) is the focal length of the eyepiece.In this problem, using the image height of approximately -1.9 cm and the focal length of the eyepiece, 15.0 cm, the angular size is calculated to be approximately 0.127 radians. This provides a tangible sense of how large the final image appears when viewed through the eyepiece, giving an observer an impression of its scale compared to direct observation.

Focal Length Calculation

The focal length of a lens is a fundamental property that determines how strongly it converges or diverges light. In the context of telescopes, focal length calculations allow us to understand the optical characteristics of the system. Calculating focal length often involves using the lens formula or applying specific scenarios, such as when dealing with distant objects where image formation occurs at the focal point. The more extended focal length of the objective lens, 95.0 cm, compared to the eyepiece's 15.0 cm, highlights how this setup is optimized for magnifying distant objects. The focal length relates directly to the telescope's ability to gather light and resolve detail.
The longer the focal length of the objective, the more significant the potential magnification, and the better the telescope can reveal fine details of the object under observation. Understanding focal length is crucial to optimizing a telescope's performance for specific observational needs.