A diverging lens (f1 = −11.0 cm) is located 51.0 cm to the left of a converging lens (f2 = 30.5 cm). A 2.5-cm-tall object stands to the left of the diverging lens, exactly at its focal point. Determine the distance of the final image relative to the converging lens.
Image by diverging lens:
object distance, u = -11 cm, focal length f1 = -11 cm
Applying thin lens equation:
1/v - 1/u = 1/f1
1/v - 1/(-11) = 1/(-11)
solving, we get, v = -5.5 cm
Now, image from converging lens:
object distance, u = -(51 + 5.5) = -56.5 cm
focal length, f2 = 30.5 cm
Applying thin lens equation:
1/v - 1/u = 1/f2
1/v - 1/(-56.5) = 1/30.5
solving, we get, v = 66.3 cm
final image is located 66.3 cm behind the converging lens
A diverging lens (f1 = −11.0 cm) is located 51.0 cm to the left of a...
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A diverging lens (f1 = −11.5 cm) is located 25.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0-cm-tall object stands to the left of the diverging lens, exactly at its focal point. What is the height of the final image (including proper algebraic sign)?
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