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A telescope is constructed from two lenses with focal lengths of \(95.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm},\) the \(95.0 \mathrm{~cm}\) 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?

Short Answer

Expert verified
The angular magnification of the telescope is -6.33, meaning the image is inverted. The height of the image formed by the objective of a building 60.0 m tall and 3.00 km away is 1.9 cm. The angular size of the final image as viewed by an eye very close to the eyepiece is 7.27°.

Step by step solution

01

Compute Angular Magnification

Angular magnification \(M\) of a telescope can be calculated using the formula \(M = -f_o/f_e\), where \(f_o\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece lens. Using the given values, \(f_o = 95.0 cm\) and \(f_e = 15.0 cm\), we substitute these into the formula to get \(M = -(95.0 cm) / (15.0 cm) = -6.33\). The negative sign means that the image is inverted, which is typical for astronomical telescopes.
02

Compute Height of Image Formed by the Objective

The height \(h\) of the image formed by the objective can be calculated using the formula \(h = f_o * h_o / d\), where \(h_o\) is the height of the object and \(d\) is the distance to the object. Using the given values, \(h_o = 60.0 m = 6000 cm\) (converted to centimeters because our focal lengths are given in cm) and \(d = 3.0 km = 300000 cm\) (again, converted to cm), we substitute these into the formula to get \(h = (95.0 cm * 6000 cm) / 300000 cm = 1.9 cm\).
03

Compute the Angular Size of the Final Image

The angular size \(θ\) of the final image can be calculated using the formula \(θ = h / f_e\), where \(h\) is the height of the image formed by the objective and \(f_e\) is the focal length of the eyepiece. Using the given values, \(h = 1.9 cm\) and \(f_e = 15 cm\), we substitute these into the formula to get \(θ = (1.9 cm) / (15 cm) = 0.127\) radians, or the equivalent in degrees using the conversion \(1 radian = 57.3 degrees\), which gives \(θ = 0.127 * 57.3 = 7.27°\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Focal Length
In the world of telescopes, understanding the focal length is essential. Focal length, denoted as \(f\), is the distance between the lens and the point where parallel rays of light converge to a focus. This measurement is crucial because it influences the magnification and clarity of the image your telescope will produce.
The focal length of a lens determines how much it will magnify objects. In this scenario, we have two lenses with focal lengths of 95.0 cm and 15.0 cm. The 95.0 cm lens is used as the objective lens, which is the larger lens at the front of the telescope. This lens collects light and focuses it to form an image.
The eyepiece, with a focal length of 15.0 cm, is the lens you look through. It magnifies the image formed by the objective lens. This combination of lenses helps in creating a detailed view of distant objects like stars and planets.
Understanding focal lengths also helps in calculating the angular magnification of the telescope, which is determined by the ratio of the focal lengths of the objective and eyepiece lenses.
Image Formation
The formation of an image in a telescope involves capturing light rays from distant objects and focusing them to create a visual representation. This process begins with the objective lens. In our case, it has a focal length of 95.0 cm. This lens gathers light from the object, like a building, and brings it to a focus, forming a real, inverted image.
In telescopes, when we say that both the object being viewed and the final image are at infinity, it means the light rays entering the telescope are almost parallel. This setup is optimal for viewing celestial objects, which are far away. The real image formed by the objective lens for a building of 60.0 meters, located 3.00 km away, can be calculated. Converting these measurements to the same units, the objective lens forms an image 1.9 cm tall. This small yet detailed image is then magnified by the eyepiece.
Angular Size
Angular size is a crucial concept when talking about telescopes and how we perceive distant objects. It refers to the perceived size of an object as seen from a particular point and is usually measured in degrees or radians. Angular size depends on both the actual size of the object and its distance from the observer.
In our example, after the objective lens forms an image of the building, we determine the angular size of this image when viewed through the telescope's eyepiece. Using the height of the image, 1.9 cm, and the focal length of the eyepiece, 15.0 cm, we can calculate the angular size. The formula for angular size \(θ\) is \(θ = h / f_e\), where \(h\) is the image height. Substituting in our values gives an angular size of 0.127 radians.
This angular size can also be converted to degrees using the conversion \(1 \text{ radian} = 57.3 \text{ degrees}\), resulting in approximately 7.27°. This measurement helps astronomers understand the scale of what they’re observing through the telescope.

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Most popular questions from this chapter

A converging lens with a focal length of \(9.00 \mathrm{~cm}\) forms an image of a \(4.00-\mathrm{mm}\) -tall real object that is to the left of the lens. The image is \(1.30 \mathrm{~cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

A compound microscope has an objective lens with focal length \(14.0 \mathrm{~mm}\) and an eyepiece with focal length \(20.0 \mathrm{~mm}\). The final image is at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) What is the distance between the two lenses? (b) Without making the approximation \(s_{1} \approx f_{1},\) use \(M=m_{1} M_{2}\) with \(m_{1}=-s_{1}^{\prime} / s_{1}\) to find the overall angular magnification of the microscope. (c) What is the percentage difference between your result and the result obtained if the approximation \(s_{1} \approx f_{1}\) is used to find \(M ?\)

To determine the focal length \(f\) of a converging thin lens, you place a \(4.00-\mathrm{mm}\) -tall object a distance \(s\) to the left of the lens and measure the height \(h^{\prime}\) of the real image that is formed to the right of the lens. You repeat this process for several values of \(s\) that produce a real image. After graphing your results as \(1 / h^{\prime}\) versus \(s\), both in \(\mathrm{cm}\), you find that they lie close to a straight line that has slope \(0.208 \mathrm{~cm}^{-2}\). What is the focal length of the lens?

You hold a spherical salad bowl \(60 \mathrm{~cm}\) in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a \(35 \mathrm{~cm}\) radius of curvature. (a) Where is the image of your \(5.0-\mathrm{cm}\) -tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of \(18.0 \mathrm{~cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\mathrm{cm}\) -tall coin that is \(6.00 \mathrm{~cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

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