Question:
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 \(10^{-6}m^2\)).
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
- 100
- 200
- 300
- 400
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The Correct Option is C
Solution and Explanation
Concept:
\[
E = \frac{hc}{\lambda}, \quad N = \frac{IA}{E}
\]
Step 1: Energy per photon \[ E = \frac{6.6\times10^{-34}\cdot3\times10^8}{5.6\times10^{-7}} \approx 3.5\times10^{-19}J \]
Step 2: Power entering eye \[ P = IA = 10^{-10}\times10^{-6} = 10^{-16}W \]
Step 3: Number of photons \[ N = \frac{10^{-16}}{3.5\times10^{-19}} = 300 \]
Step 1: Energy per photon \[ E = \frac{6.6\times10^{-34}\cdot3\times10^8}{5.6\times10^{-7}} \approx 3.5\times10^{-19}J \]
Step 2: Power entering eye \[ P = IA = 10^{-10}\times10^{-6} = 10^{-16}W \]
Step 3: Number of photons \[ N = \frac{10^{-16}}{3.5\times10^{-19}} = 300 \]
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