An equiconvex lens of focal length 10cm (at t=0) and refractive index (1.5) is placed in a medium whose refractive index varies with time as μ=1+t10; t in seconds. The value of t when the lens acts as a diverging lens of focal length 20cm is

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answer is B.

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Detailed Solution

−120=1.5μ−12R and 110=1.5−12R=1R⇒∴R=10cmSo, μ=2∴t=10 secs

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