A 1.00 cm high object is placed 3.00 cm to the left of a converging lens of foca

Question

A 1.00 cm high object is placed 3.00 cm to the left of a converging lens of focal length 7.40 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.

The position of the final image is ___ cm ___________ (in front of the second lens / behind the second lens / in front of the first lens), and its height is ___ cm.

Is the image inverted or upright? Real or virtual?
- virtual and inverted
- real and upright
- real and inverted
- virtual and upright

Explanation / Answer

1/f = 1/v - 1/u 1/7.4 = 1/v + 1/3 v = - 5.045 cm u for 2 lens = 6 + 5.045 = 11.045 cm 1/-16 = 1/v + 1/11.045 v = - 6.534 cm final position is 6.534 cm front of the second lens. b) m1 = v/u = 5.045 /3 = 1.6816 m2 = v/u = 6.534/11.045 = 0.5915 H = 1 x m1 m2 = 0.9948 cm virtual and upright

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