
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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A converging lens with a focal length of 12.8 cm forms a virtual image 8.50 mm tall, 17.4 cm to the right of the lens.
What is the position and size of the object?
Will the image be inverted or upright?
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- An object is located 16.6 cm to the left of a diverging lens having a focal length f = −32.4 cm. What is the magnification of the image? Group of answer choices 0.661 2.13 0.346 1.17arrow_forwardCan you help me calculate the height. I kept getting it wrong.arrow_forwardA 1.00-cm-high object is placed 4.30 cm to the left of a converging lens of focal length 8.45 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. position height cm ---Select--- cmarrow_forward
- It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance ss to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance s′s′ and then use the equation 1s+1s′=1f1s+1s′=1f, to calculate the focal length ff of the lens. But this procedure won't work with a diverging lens−−by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object 20.0 cmcm to the left of the lens, the image is 29.7 cmcm to the right of the lens.…arrow_forwardAn object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.140 m in front of a diverging lens. The focal length of the lens is -0.25 m. (Include the sign of the value in your answers.) (a) What is the magnification? (b) What is the image height? m (c) What is the image distance? marrow_forwardA 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 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. position 7.5 cm in front of the second lens v 0.5466 height Calculate 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. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? real virtualarrow_forward
- I continue to get the wrong answer. TIA.arrow_forwardA 3 cm tall object is placed 2 cm to the left of a converging lens that has a focal length with a magnitude of 4 cm. A diverging lens with a focal length of magnitude 8 cm is placed 10 cm to the right of the first lens. What is the magnification of the final image produced by these two lenses? Make sure you say whether it is bigger/smaller and upright/inverted.arrow_forwardA farsighted person can read printing as close as 25.0 cm when she wears contacts that have a focal length of 38.0 cm. One day, however, she forgets her contacts and uses a magnifying glass, as in the drawing. It has a maximum angular magnification of 7.42 for a young person with a normal near point of 25.0 cm. What is the maximum angular magnification that the magnifying glass can provide for her? Virtual Magnifying glass Object LE ho dj. imagearrow_forward
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