Question:medium

Minimum intensity of light detectable is \(10^{-10}\,W/m^2\). Number of photons entering eye per second (\(\lambda=5.6\times10^{-7}m\), pupil area \(10^{-6}m^2\)).

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Number of photons = total energy per second energy per photon.
Updated On: Apr 23, 2026
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The Correct Option is C

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