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Question

l htps:lsession.masteringphysics.com/myct/itemView?assignmentProblemID-89973445 Constants Part A A converging lens with a focal length of 6.80 cm forms an image of a 4.80 mm -tall real object that is to the left of the lens. The image is 2.20 cm tall and erect Where are the object and image located? Enter your answer as two numbers separated with a comma. cIn Previous Answers Request Answer X Incorrect, Try Again; 19 attempts remaining Part B s the image real or virtual? The image is virtual Previous Answers Correct

Explanation / Answer

Given,

f = 6.8 cm ; h = 4.8 mm ; h' = 2.2 cm = 22 mm

since the image is erect, magnification mmust be positive and image distance should be negative

M = -i/o = h'/h

-(-i/o) = 22/4.8 = 4.583

i = -4.58 o

we know from lens eqn

1/f = 1/i + 1/o

1/6.8 = 1/-4.58o + 1/o = 1/o (1/4.583 + 1)

o = 5.32 cm

i = -4.58 x 5.32 = -24.34 cm

Hence, s,s' = 5.32, -24.34

B)The image is virtual

Since the image distance is negative its virtual.

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