A doctor uses a 17.5 cm focal length magnifying glass to examine a suspected skin lesion on a patient. The magnifying glass is held a distance 12.0 cm from the lesion.   (a)Calculate the position of the image of the lesion (in cm). (Include the sign of the value in your answer.)  cm   (b) Calculation the magnification of the image of the lesion. (Include the sign of the value in your answer.   (c)Calculate the height of the image of the lesion (in mm). The measured height of the lesion is 1.95 mm. mm   (d) Describe the image Inverted and smaller than the object. Inverted and larger than the object.     Upright and smaller than the object. Upright and larger than the object. (e) Determine the position of the image relative to the lens  The image is on the same side of the lens as the lesion. The image is on the opposite side of the lens as the lesion.     The image is exactly at the position of the lens. It is impossible to tell where the position of the image is relative to the lens.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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A doctor uses a 17.5 cm focal length magnifying glass to examine a suspected skin lesion on a patient. The magnifying glass is held a distance 12.0 cm from the lesion.
 
(a)Calculate the position of the image of the lesion (in cm). (Include the sign of the value in your answer.)  cm
 
(b) Calculation the magnification of the image of the lesion. (Include the sign of the value in your answer.
 
(c)Calculate the height of the image of the lesion (in mm). The measured height of the lesion is 1.95 mm. mm
 
(d) Describe the image
Inverted and smaller than the object.
Inverted and larger than the object.    
Upright and smaller than the object.
Upright and larger than the object.

(e) Determine the position of the image relative to the lens
 The image is on the same side of the lens as the lesion.
The image is on the opposite side of the lens as the lesion.    
The image is exactly at the position of the lens.
It is impossible to tell where the position of the image is relative to the lens.

 

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