A spherical mirror forms a magnified (m>1) and upright image of an 5ポイント object. To decrease the magnification, should the object be moved closer to the mirror or farther away?, and is this mirror concave or convex? * closer, convex farther away, concave closer, concave further away, convex

A spherical mirror forms a magnified (m>1) and upright image of an 5ポイント object. To decrease the magnification, should the object be moved closer to the mirror or farther away?, and is this mirror concave or convex? * closer, convex farther away, concave closer, concave further away, convex

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter26: Image Formation By Mirrors And Lenses

Section: Chapter Questions

Problem 16P: A concave mirror has a radius of curvature of 60.0 cm. Calculate the image position and...

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The lens and mirror in Figure P36.77 are separated by d = 1.00 m and have focal lengths of +80.0 cm and -50.0 cm, respectively. An object is placed p = 1.00 m to the left of the lens as shown, (a) Locate the final image, formed by light that has gone through the lens twice. (b) Determine the overall magnification of the image and (c) state whether the image is upright or inverted.
Suppose you want to use a converging lens to project the image of two trees onto a screen. As show n in Figure CQ36.9, one tree is a distance x from the lens and the other is at 2x. You adjust the screen so that the near tree is in locus. It you now want the far tree to be in focus, do you move the screen toward or away from the lens?
What are the differences between real and virtual images? How can you tell (by looking) whether an image formed by a single lens or mirror is real or virtual?
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An individual is nearsighted; his near point is 18.0 cm and his far point is 50.0 cm. (a) What lens power is needed to correct his nearsightedness? (b) When the lenses are in use, what is this persons near point?
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A microscope has an objective lens with a focal length of 16.22 mm and an eyepiece with a focal length of 9.50 mm. With the length of the barrel set at 29.0 cm, the diameter of a red blood cells image subtends an angle of 1.43 mrad with the eye. It the final image distance is 29.0 cm from the eyepiece, what is the actual diameter of the red blood cell? Hint: To solve this question, go back to basics and use the thin-lens equation.
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