Question

a 5.5 D lens is placed 14cm from an object 4mm in height. what are the position, type height of the image?

lens eq: 1/Do + 1/Di = 1/f

Answer 1

Do = .14m
Ho= .004m
Di=?
f = ?

Answer 2

f = 1/-5.5 = 18m?

Answer 3

I am having trouble with using this equation to solve for Di

Answer 4

im missing something in the algebra steps I think because I can't get the right answer, apparently its .079m = Di

Answer 5

I'm assuming you're working with a converging lens.


f = 1/5.5 = 0.1818


1/Do + 1/Di = 1/f


1/0.14 + 1/Di = 1/0.1818


7.142857 + 1/Di = 5.5


1/Di = 5.5 - 7.142857


1/Di = -1.642857


Di = 1/(-1.642857)


Di = -0.6086957


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Magnitude:

M = -Di/Do

M = -(-0.6086957/0.14)

M = 4.347826

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M = Hi/Ho

4.347826 = Hi/0.004

4.347826*0.004 = Hi

0.017391304 = Hi

Hi = 0.017391304


Since the image height Hi is positive, this means that the image is right side up (eg if the object is upright, then the image is upright).


Since the image distance Di is negative, this means that the image is a virtual image.