A converging lens with a focal length of 13.5 cm and a diverging lens are placed

Question

A converging lens with a focal length of 13.5 cm and a diverging lens are placed 29.0 cm apart, with the converging lens on the left. A 2.00-cm high object is placed 22.0 cm to the left of the converging lens. The final image is 34.0 cm to the left of the converging lens. (a) What is the focal length of the diverging lens? Incorrect: Your answer is incorrect. cm (b) What is the height of the final image? Incorrect: Your answer is incorrect. cm (c) Is the final image upright or inverted? upright inverted Correct: Your answer is correct.

Explanation / Answer

u1 = 0.22 m: f1 = 0.135 m: sep = 0.29 m: v2 = -(sep + 0.34) m = -0.63 m
For the 1st lens:
v1 = 1 / (1/f1 -1/u1) = 0.35 m
For the 2nd lens:
u2 = sep - v1 = -0.06 m (the intermediate image is to the right of lens 2)
v2 = 1 / (1/f2 - 1/u2), but since we want f2, we write
f2 = 1 / (1/v2 + 1/u2)
This can be solved since we know v2 = -(sep + 0.34) = -0.63 m.
A. The answer is f2 = -0.053 m
B. The magnification is v1v2/(u1u2) = 16.7
C. + magnification = upright image.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.