You will begin with a relatively standard calculation. Consider a concave spherical mirror with a radius of curvature equal to centimeters. An object centimeters tall is placed along the axis of the mirror, centimeters from the mirror. You are to find the location and height of the image.
Part A
What is the focal length of this mirror?
Express your answer in centimeters to three significant figures.
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=30.00 Correct
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Part B
Now use the spherical mirror equation to find the image distance .
Express your answer in centimeters, as a fraction or to three significant figures.
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=90.00 Correct
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Part C
Find the magnification , using and .
Express your answer as a fraction or to three significant figures.
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=−2.000Correct
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Part D
Finally, use the magnification to find the height of the image.
Express your answer in centimeters, as a fraction or to three significant figures.
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=−12.00 Correct
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Part E
Look at the signs of your answers to determine which of the following describes the image formed by this mirror:
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real and inverted
Correct
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Next, you will manipulate the spherical mirror equation to explore the qualities (real/virtual and upright/inverted) of some images.
Part F
Solve the spherical mirror equation for .
Express your answer in terms of and .
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=(s_ob*f)/(s_ob  f)Correct
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Part G
What is the magnification ? Use your answer from Part F.
Express your answer in terms of and .
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=(1/((1/f)(1/s_ob)))/s_obCorrect
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Part H
If you are working with a convex mirror (), which of the following describes the image?
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virtual and upright
Correct
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You know now that the magnification of a single convex mirror is always positive.
Part I
What can one say about the size of the magnification of a single convex mirror?
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It is less than one.Correct
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Because they create an upright image that is smaller than the object, convex mirrors are used in many different places. You will see them in convenience stores for surveillance, because they allow one person to see most of the store at a glance. They are also used in hospitals, parking garages, and winding roads to provide a limited view around a corner to aid in avoiding collisions.
As the radius of curvature of a spherical mirror gets larger, the mirror becomes flatter. Notice that as goes to infinity, so does , because . Thus, as gets larger, gets smaller. In the limit where you allow to go to infinity, becomes zero. Therefore, if you could construct a mirror with an infinitely large radius of curvature, it would obey the equation .
Part J
What is the value of obtained from this new equation?
Express your answer in terms of .
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=s_obCorrect
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The introduction to this part noted that as gets larger, the mirror gets flatter. Recall that the earth is approximately a sphere. It only appears flat to us because of its large radius of curvature. From this, you should be able to see that you can in fact make a mirror with infinite radius of curvature. It would be a flat mirror! Your answer to Part J is the equation relating and for a plane mirror.
Understanding Multiple Optics
In order to find the location of the final image of the object formed by this system, you will need to trace the rays through the system, instrument by instrument. You are strongly advised to draw a picture with the x axis and the location of the lens, mirror, and object marked. Then, as you proceed through the problem, you can mark where each image is located.
Part A
First, find the location of the image created by the lens by itself (as if no other instruments were present).
Express your answer in meters, to three significant figures or as a fraction.
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=2.500 Correct
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Part B
Next, find the location of the image created by the plane mirror (after the light passes through the lens).
Express your answer in meters, to three significant figures or as a fraction.
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=−2.500 Correct
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Proceed with caution! The light reflects off of the mirror and back through the lens a second time!
Part C
What is the location of the final image, as seen by an observer looking toward the mirror, through the lens? Keep in mind that the light must pass back through the lens, and thus you must do one more calculation with the thin lens equation.
Express your answer in meters, to three significant figures or as a fraction.
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=20.00 Correct
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Part D
Is the final image formed by this system real or virtual?
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real
Correct
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If you have more than one optical instrument, the total magnification is equal to the product of the individual magnifications. Keep in mind that the image of one device becomes the object of the next. If the first device creates an image that is ten times the size of the object, and the second creates an image that is twenty times the size of the first image (which was its object), then the final image will be two hundred times the size of the original object. This makes sense because the second device magnifies further what was already magnified by the first device. To find the magnitude of the magnification of the final image, you will need to consider each instrument and find the individual magnifications.
Part E
First, find the magnitude of the magnification of the image created when light from the object passes through the lens the first time (as if the mirror were not present).
Express your answer to three significant figures or as a fraction.
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=1.000Correct
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Part F
Next, find the magnitude of the magnification of the plane mirror.
Express your answer to three significant figures or as a fraction.
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=1.000Correct
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Of course, this is expected. Plane mirrors don't magnify your image!
Part G
Now find the magnitude of the magnification of the image created when light from the object passes through the lens the second time (after reflecting off of the mirror).
Express your answer to three significant figures or as a fraction.
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=0.5000Correct
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Part H
What is the magnitude of the magnification of the final image?
Express your answer to three significant figures or as a fraction.
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=0.5000Correct
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Part I
Is the final image upright or inverted? Give the orientation relative to the original object.
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upright
Correct
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