(a) A 1.00 cm-high object is placed 3.15 cm to the left of a converging lens of focal length 7.95 cm. A diverging lens of focal length -16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image.positioncm in front of the second lens ✔height6.63.07xCalculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cmIs the image inverted or upright?ⒸuprightO invertedIs the image real or virtual?O realO virtual(b) What If? If an image of opposite characteristic, i.e., virtual if the image in part (a) is real and real if the image in part (a) is virtual, is to be obtained, what is the minimum distance (in cm), and in which direction, that the object must be moved from its originalposition?distancedirection10.15to the left✔X cm

Given : Height of the object, h = 1 cm Object distance, u = 3.15 cm Focal length of…

Answered: (a) A 1.00 cm-high object is placed…

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 70P

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Similar questions

The power for normal close vision is 54.0 D. In a vision-correction procedure, the power of a patient’s eye is increased by 3.00 D. Assuming that this produces normal close vision, what was the patient’s near point before the procedure?
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The contact-lens prescription for a nearsighted person is —4.00 D and the person has a far point of 22.5 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?
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Two thin lenses of focal lengths f1 = 15.0 and f2 = 10.0 cm, respectively, are separated by 35.0 cm along a common axis. The f1 lens is located to the left of the f2 lens. An object is now placed 50.0 cm to the left of the f1 lens, and a final image due to light passing though both lenses forms. By what factor is the final image different in size from the object? (a) 0.600 (b) 1.20 (c) 2.40 (d) 3.60 (e) none of those answers
Show that the magnification of a thin lens is given by M = di/do (Eq. 38.6). Hint: Follow the derivation of the lens makers equation (page 1233) and start with a thick lens.
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Two stars that are 109km apart are viewed by a telescope and found to be separated by an angle of 105 radians. If the eyepiece of the telescope has a focal length of 1.5 cm and the objective has a focal length of 3 meters, how far away are the stars from the observer?
A converging lens made of crown glass has a focal length of 15.0 cm when used in air. If the lens is immersed in water, what is its focal length? (a) negative (b) less than 15.0 cm (c) equal to 15.0 cm (d) greater than 15.0 cm (e) none of those answers

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