An object is placed on an optical bench. A lens with a focal length of 15 cm is

Question

An object is placed on an optical bench. A lens with a focal length of 15 cm is placed 30 cm from the object. A diverging lens with a focal length of -10 cm is placed 7.0 cm beyond that. Where is the final image formed, relative to the second lens?

An object is placed on an optical bench. A lens with a focal length of 15 cm is placed 30 cm from the object. A diverging lens with a focal length of -10 cm is placed 7.0 cm beyond that. Where is the final image formed, relative to the second lens?

17 cm beyond the diverging lens 20 cm behind the diverging lens 18 cm behind the diverging lens 30 cm beyond the diverging lens 45 cm beyond the diverging lens

Explanation / Answer

For the first lens, apply 1/f = 1/p + 1/q

1/15 = 1/30 + 1/q

q = 30 cm

For the second lens, the object distance is 7 - 30 = -23 cm

1/-10 = 1/-23 + 1/q

q = -17.6

That means the final image is 17.6 cm, which rounds to 18 cm, behind the diverging lens. The third choice in your list.

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