A diverging lens has a focal length of -7.20 cm A) What are the image distances

Question

A diverging lens has a focal length of -7.20 cm

A) What are the image distances q for objects placed at the distances p from the lens which are tabulated below? In each case, calculate the magnification m (I ALREADY DID THIS PART-BELOW ARE THE ANSWERS, I NEED HELP WITH PART B....)

1) Object distance (cm): 5.00, Image distance (cm): -2.95, magnification: .59

2) Object distance: 7.20, Image distance: -3.6, magnification: .5

3) Object distance: 14.0, Image distance: -4.75, magnification: .34

4) Object distance: 16.0, Image distance: -4.97, magnification: .31

5) Object distance: 20.0, Image distance: -5.29, magnification: .26

B) *****PLEASE HELP WITH THIS PART*****

If the object is 4.20 cm high, what is the height of the image for the object distances of 5.00 cm and 20.0 cm?

Explanation / Answer

1/v - 1/u =1/f
here, f = -40, u = -24
:. 1/v = -(1/24 +1/40) = -1/15
So, the image will be formed at 15cm on the same side of the lens as that of the object. It will be virtual.
Magnification(m) = v/u = 15/24 = 0.625

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