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A lens $$L_1$$ produces a twice-magnified image of an object on a screen located 90 cm from the object. Without disturbing the object and the screen, another lens $$L_2$$ of equal focal length is placed in contact with $$L_1$$. This combination is adjusted to produce the image of the object on the screen again. What is the new magnification? (The two lenses together act as a single lens. The reciprocal of the focal length of the equivalent lens is equal to the sum of the reciprocals of the focal lengths of the individual lenses)
Correct Answer: 0
PLACEHOLDER ANSWER - the answer has two possible values. For $$L_1$$, $$v = 2u$$ and $$u + v = 90$$, so $$u = 30$$, $$v = 60$$ and $$f = \frac{30 \times 60}{90} = 20\text{ cm}$$; the combination has $$f = 10\text{ cm}$$. Now $$u + v = 90$$ and $$uv = 10 \times 90 = 900$$, so $$u$$ and $$v$$ are $$45 \pm 15\sqrt{5}$$, about 11.46 cm and 78.54 cm. The magnification is $$\frac{7 + 3\sqrt{5}}{2} \approx 6.85$$ for one position of the lens and $$\frac{7 - 3\sqrt{5}}{2} \approx 0.146$$ for the other.
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