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The diameter of Mars is 6794 km, and its minimum distance from the earth is 5.58 \(\times\) 10\(^7\)km. When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.75 m.

Short Answer

Expert verified
The diameter of the image of Mars is approximately 0.00000000214 m.

Step by step solution

01

Understand the Problem

We need to find the diameter of the image of Mars as seen through a telescope. The problem gives us the diameter of Mars and its minimum distance from Earth, and also the focal length of the telescope. We will use these values to calculate the image diameter.
02

Use the Lens Formula

First, recognize that we will use the mirror formula for a concave mirror, which similar to the lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \). Here, \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Convert the focal length from meters to kilometers to match the units. Thus, 1.75 m = 0.00175 km.
03

Calculate Object Distance

The object distance \( u \) is the distance from Mars to Earth, which is given as \( 5.58 \times 10^7 \) km. Substitute this value into the mirror formula.
04

Solve for Image Distance

Substitute the known values into the mirror equation: \( \frac{1}{0.00175} = \frac{1}{v} - \frac{1}{5.58 \times 10^7} \). Solve this equation for \( v \), the image distance.
05

Use Magnification Formula

Once \( v \) is found, use the magnification formula: \( m = \frac{h_i}{h_o} = \frac{v}{u} \), where \( h_i \) is the image height and \( h_o \) is the object height (diameter of Mars). Thus, \( h_i = h_o \times \frac{v}{u} \).
06

Substitute Mars' Diameter and Calculate Image Diameter

Substitute \( h_o = 6794 \) km and the computed \( v \) into the magnification formula to find \( h_i \), the image diameter. Convert this diameter to meters, since the focal length was initially given in meters.

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Key Concepts

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

Concave Mirror
A concave mirror is a type of mirror that is curved inward, resembling a portion of the interior of a sphere. These mirrors are capable of focusing light to a specific point. This unique property is what makes them useful in telescopes and other optical instruments.

The way concave mirrors work is based on the principle of reflection. When light rays parallel to the principal axis (an imaginary line that splits the mirror in half) hit the surface of a concave mirror, they reflect back through the focus. This focus is a point along the principal axis where these reflected rays meet or converge.
  • Used to form real or virtual images, depending on the position of the object.
  • Commonly found in devices like telescopes, headlights, and shaving mirrors.
The focal length of a concave mirror, the distance from the mirror to the focus, determines its strength to magnify or minimize an image. In the given exercise, the concave mirror in the telescope helps form an image of Mars by collecting and focusing light rays from the planet.
Image Formation
In telescope optics, image formation involves using mirrors or lenses to visually present distant objects in a perceivable format. For concave mirrors, image formation relies on the mirror formula, given as: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Here, \(f\) is the focal length, \(v\) is the image distance (the distance from the mirror to where the image is formed), and \(u\) is the object distance (how far the object is from the mirror).

In the exercise, Mars is the object; thus, the formula is used to determine where the image of Mars forms once it reflects off the mirror. We must ensure units are consistent—for example, converting the focal length from meters to kilometers.

For most telescopes utilizing concave mirrors:
  • The image can be real (formed on the same side as the object) or virtual (on the opposite side, behind the mirror).
  • The clear, sharp image depends on accurately calculating the image distance.
This mathematical approach helps us locate where the stark and recognizable image of celestial bodies will appear through the concave mirror.
Magnification Formula
The magnification formula is a helpful tool in determining how much larger or smaller the image will appear in comparison to the actual object. With concave mirrors, the formula is:\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]Where \( h_i \) is the image height, \( h_o \) is the object height, \( v \) is the image distance, and \( u \) is the object distance.

This formula states that the magnification (\( m \)) of the image is proportional to the ratio of image distance to object distance. By rearranging to find \( h_i \), we can determine the image diameter of Mars once calculated.
  • If \( m > 1 \), the image is larger than the object.
  • If \( m < 1 \), the image is smaller.
  • If \( m = 1 \), the image is the same size as the object.
In this specific exercise, substituting the given planet diameter and distances helps calculate the diameter of Mars' image as it would appear through the telescope's concave mirror. Understanding these relationships is crucial to predicting and interpreting how objects from space can be represented visually.

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Most popular questions from this chapter

The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.20 m, and the final image is at infinity. What is the angular magnification of the telescope?

A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image \(without\) a lens. (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (\(Hint\): Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is 25 cm square and 20.0 cm deep, with the hole in the middle of one of the 25 cm \(\times\) 25 cm faces. If this camera is used to photograph a fierce chicken that is 18 cm high and 1.5 m in front of the camera, how large is the image of this bird on the film? What is the lateral magnification of this camera?

A converging lens with a focal length of 70.0 cm forms an image of a 3.20-cm- tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and \(2\over12\) times the size of the object? Where is the image?

(a) For a lens with focal length f, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)

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