A 1.00-cm-high object is placed 4.25 cm to the left of a converging lens of foca

Question

A 1.00-cm-high object is placed 4.25 cm to the left of a converging lens of focal length 7.60 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 Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm 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 inverted Is the image real or virtual? real virtual

Explanation / Answer

A)

1/f = 1/v - 1/u

1/7.6 = 1/v + 1/4.25

0.13158 = 1/v + 0.23529

v = 1/-0.10371

v = -9.642 cm

u for 2 lens = 6 + 9.642 = 15.642 cm

1/-16 = 1/v + 1/15.642

-0.0625 = 1/v + 0.0639

v = 1/-0.1264

v = -7.911 cm

final position is 7.91 cm front of the second lens.

B)

m1 = v/u = 9.642 /4.5 = 2.143

m2 = v/u = 7.911/15.642 = 0.506

H = 1*m1*m2

= 1*2.143*0.506

= 1.084 cm

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