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Chapter 24: Problem 7

The diameter of Mars is \(6794 \mathrm{~km}\), and its minimum distance from the earth is \(5.58 \times 10^{7} \mathrm{~km} .\) (a) 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 \(1.75 \mathrm{~m}\). (b) Where is the image located?

Short Answer

Expert verified
The image diameter is about 0.000213 m, and the image is located approximately 1.75 m from the mirror.

Step by step solution

01

Write Down the Given Data

We know the following information about the problem: \( D_{\text{Mars}} = 6794 \text{ km} \), the minimum distance from Earth \( d_e = 5.58 \times 10^7 \text{ km} \), and the focal length of the telescope \( f = 1.75 \text{ m} \). These will be used in the calculations.
02

Convert Units

Convert the diameter of Mars from kilometers to meters since the focal length of the mirror is in meters. Thus \( D_{\text{Mars}} = 6794 \times 10^3 \text{ m} \). Similarly, convert the minimum distance from Earth to meters: \( d_e = 5.58 \times 10^7 \times 10^3 \text{ m} \).
03

Use the Lens Formula to Find Image Distance

We use the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance and \( d_i \) is the image distance. Plug in \( f = 1.75 \text{ m} \) and \( d_o = d_e \):\[ \frac{1}{1.75} = \frac{1}{5.58 \times 10^{10}} + \frac{1}{d_i} \]Solving this equation, \( \frac{1}{d_i} = \frac{1}{1.75} - \frac{1}{5.58 \times 10^{10}} \). Calculate \( d_i \).
04

Calculate the Image Diameter

Use the magnification formula \( m = \frac{d_i}{d_o} = \frac{D_i}{D_o} \) to calculate the image diameter \( D_i \). Given \( D_o = 6794 \times 10^3 \text{ m} \) and \( d_o = 5.58 \times 10^{10} \text{ m} \), and \( d_i \) from the previous step:\[ m = \frac{d_i}{5.58 \times 10^{10}} \]\[ D_i = m \times D_o \]Calculate \( D_i \).
05

Determine Image Location

From the mirror equation and the calculations done in Step 3, since the focal length is 1.75 m and \( d_i \) is calculated, the image will be located at distance \( d_i \) from the mirror. Check if \( d_i \) is positive (real image) or negative (virtual image).

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Key Concepts

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

Understanding Concave Mirrors
Concave mirrors are a type of mirror that is curved inward, resembling a portion of the inside of a sphere. This curvature causes parallel rays of light to converge at a single point called the focal point.
These mirrors are often used in telescopes due to their ability to focus light efficiently.
  • Optical Axis: The central line that runs perpendicular to the mirror's surface.
  • Focal Point (F): The point at which light rays parallel to the optical axis converge after reflecting off the mirror.
  • Center of Curvature (C): The center of the sphere from which the mirror is derived. It's twice the focal length from the mirror.
These concepts allow concave mirrors to form real or virtual images depending on the object's position relative to the focal point. The ability of these mirrors to generate magnified images finds utility in astronomical telescopes to study celestial objects.
Image Formation with Concave Mirrors
Image formation in concave mirrors is governed by the mirror equation and ray diagrams. When an object is placed at a certain distance from a concave mirror, the characteristics of the image formed depend on the object's position in relation to the mirror's focal point (F) and center of curvature (C).
  • Beyond C: The image is real, inverted, and smaller than the object.
  • At C: The image is real, inverted, and the same size as the object.
  • Between F and C: The image is real, inverted, and larger than the object.
  • At F: No image is formed as rays are parallel.
  • Within F: The image is virtual, upright, and larger than the object.
Understanding these principles is crucial for astronomers who use telescopes with concave mirrors to view distant objects and to predict the type of image that will be formed.
Focal Length in Telescope Optics
The focal length is a critical parameter in telescope optics, determining how a concave mirror focuses light. It is the distance between the mirror's surface and its focal point. For a concave mirror, the focal length is generally a negative value in mirror formulas, signifying the point is in front of the mirror.
This property influences the type and characteristics of the image created:
  • Shorter Focal Length: Provides a higher degree of magnification, making minute details more visible. It also results in a wider field of view.
  • Longer Focal Length: Offers lower magnification but produces detailed images of larger objects.
In the context of telescopes, adjusting the focal length allows astronomers to switch between gaining a broader view of the sky and zooming in on specific celestial bodies.
Utilizing the Magnification Formula
The magnification formula is crucial in determining the size of the image produced by a telescope's mirror. It relates the image distance (\(d_i\)) and object distance (\(d_o\)) to the actual and perceived sizes of the object, represented as image height (\(D_i\)) and object height (\(D_o\)).
The formula is:\( m = \frac{d_i}{d_o} = \frac{D_i}{D_o} \)
  • Magnification Ratio: The ratio \( m \) tells us how many times larger or smaller the image is compared to the object.
  • Application in Telescopes: When the image distance is known, this formula helps in calculating the apparent size of astronomical bodies.
  • Real vs Virtual Images: A positive \(m\) implies a real (inverted) image, while a negative \(m\) indicates a virtual (upright) image.
Applying this formula is essential for astronomers because it transforms how they perceive distant objects via their telescopes, allowing for precise analysis of celestial dimensions.

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Most popular questions from this chapter

If you run away from a plane mirror at \(2.40 \mathrm{~m} / \mathrm{s},\) at what speed does your image move away from you?

A 3.80-mm-tall object is \(24.0 \mathrm{~cm}\) from the center of a silvered spherical glass Christmas tree ornament \(6.00 \mathrm{~cm}\) in diameter. What are the position and height of its image?

The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{~cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we can ignore it. The depth of a typical human eye is around \(25 \mathrm{~mm}\). (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were \(25 \mathrm{~cm}\) in front of the eye? If not, where would it focus that text, in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{~mm},\) where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude \(18.0 \mathrm{~cm}\). (a) Another car is seen in this side mirror and is \(13.0 \mathrm{~m}\) behind the mirror. If this car is \(1.5 \mathrm{~m}\) tall, what is the height of its image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?

Two double-convex thin lenses each have surfaces with the same radius of curvature of magnitude \(2.50 \mathrm{~cm} .\) However, lens 1 has a focal length of \(f_{1}=2.5 \mathrm{~cm}\) and lens 2 has a focal length of \(f_{2}=1.25 \mathrm{~cm} .\) Find the ratio of the indices of refraction of the two lenses, \(n_{1} / n_{2}\).
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