A screen is placed 45.0 cm to the right of the converging lens. A diverging lens

Question

A screen is placed 45.0 cm to the right of the converging lens. A diverging lens is placed between the converging lens and the screen. When the diverging lens is 9.8 cm to the right of the converging lens the image formed by the lens combination is located at the screen. What is the focal length f_2 of the diverging lens? A screen is placed a distance d = 41.5 cm to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +5.55 cm be placed if the real image of the object is at the location of the screen?

Explanation / Answer

1) need additional informatkon of converging lens

2)

s + s' = 41.5 cm

s' = 41.5 - s

for image to be real

1/s + 1/s' = 1/f

1/s + 1/(41.5-s) = 1/5.55

s = 6.6 cm

s = 35 cm

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