Problem 65 - University Physics with Modern 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 for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 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) -6.33, (b) -0.019 cm, (c) 0.1266 radians.

Step 1: Understanding Telescope Magnification

The angular magnification of a 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 lens. For this problem, use \( f_o = 95.0 \; \text{cm} \) and \( f_e = 15.0 \; \text{cm} \). Substitute the values:\[ M = -\frac{95.0}{15.0} = -6.33 \] So, the angular magnification is \(-6.33\).

Step 2: Calculating Image Height with Objective Lens

To find the height of the image formed by the objective lens, use the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( f = 95.0 \; \text{cm} \), \( d_o = 3000 \; \text{m} = 300,000 \; \text{cm} \), and we seek \( d_i \) near \( f \) since the object is far away, we assume \( \frac{1}{d_i} \approx \frac{1}{f} \).The height of the image \( h_i \) is given by the magnification formula: \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \). Since \( d_i \approx f \) and \( h_o = 60.0 \; \text{m} = 6000 \; \text{cm} \), we have:\[ h_i = -\frac{95.0}{300,000} \times 6000 \approx -0.019 \; \text{cm} \].Therefore, the height of the image is \(-0.019 \; \text{cm}\).

Step 3: Determining Angular Size of Final Image

The angular size of the final image as seen by an eye close to the eyepiece is calculated by considering the angular magnification and original angular size. The original angular size \( \theta_o \approx \frac{h_o}{d_o} = \frac{60.0}{3000} = 0.02 \), and the angular size \( \theta_i \) is:\[ \theta_i = M \times \theta_o = -6.33 \times 0.02 = -0.1266 \; \text{radians} \].The negative sign indicates the image is inverted, and the angular size is \(0.1266\) radians.

Key Concepts

Angular Magnification

Angular magnification is a key concept in understanding how telescopes work. It refers to how much larger an object appears when viewed through a telescope compared to the naked eye. Mathematically, angular magnification \( M \) is given by the formula \( M = -\frac{f_o}{f_e} \), where \( f_o \) is the focal length of the telescope's objective lens, and \( f_e \) is the focal length of the eyepiece lens. The negative sign simply indicates that the image is inverted. For our exercise, with an objective lens of 95.0 cm and an eyepiece lens of 15.0 cm, the calculation results in an angular magnification of \(-6.33\), meaning the image is 6.33 times larger and inverted compared to the view with the unaided eye.
Angular magnification helps astronomers and casual star gazers to see distant celestial bodies in greater detail than possible with the naked eye.

Objective Lens

The objective lens is a crucial component of a telescope. It is the larger lens that first captures light from a distant object and creates a real image. Its focal length determines the telescope's light-gathering power and resolution. In our problem, we have a 95.0 cm objective lens, making it responsible for creating a clean, clear initial image of the distant object.
The focal length of the objective lens affects the telescope's overall magnification and helps determine the quality and brightness of the image. The larger the objective lens, the more light it can collect, which is essential for observing dim objects like stars and galaxies.

Eyepiece Lens

The eyepiece lens in a telescope is the smaller lens closest to your eye. It further magnifies the image formed by the objective lens, allowing you to see the enlarged picture. What you see when you look through the telescope is the magnified version of the image formed by the objective lens.
Its focal length, in conjunction with that of the objective lens, determines the overall magnification of the telescope system. In our example, a 15.0 cm eyepiece lens works with the 95.0 cm objective lens, contributing to the angular magnification of \(-6.33\). This relationship illustrates how adjusting the focal lengths of the eyepiece and objective lenses can alter the magnifying power of a telescope.

Lens Formula

The lens formula is crucial for understanding how lenses in a telescope create an image. It's written as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.
In telescopes, this formula helps determine where the image will form when you know the distances involved. Because the object in our telescope example is far away, \( d_o \) is very large, allowing us to simplify the formula as \( \frac{1}{d_i} \approx \frac{1}{f} \). This simplification is handy for calculating the placement of the image formed by the objective lens.

Image Height

Image height in telescopes is determined by the magnification formula \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \), where \( h_i \) is the image height and \( h_o \) is the original height. This ratio tells us how the size of the image compares to the actual size of the object being observed.
In the exercise, by substituting known values, the image height is calculated as \(-0.019 \; \text{cm}\). This negative value indicates that the image is inverted, common in refracting telescopes. Understanding image height helps in visualizing what is captured and projected within a telescope.

Angular Size

Angular size is a measure of how large an object appears to an observer when looking through a telescope. It is affected by both the actual size of the object and its distance from the observer. In our problem, the original angular size \( \theta_o \) is calculated as \( \theta_o = \frac{h_o}{d_o} = 0.02 \; \text{radians} \).
The final angular size \( \theta_i \) seen through the telescope is affected by the angular magnification: \( \theta_i = M \times \theta_o \). For our telescope, this computes to \(-0.1266 \; \text{radians}\), indicating that the image appears larger and inverted. Angular size is an important concept for understanding how objects appear through optical instruments like telescopes.