A) 5
B) 10
C) 106
D) 15
Correct Answer: A
Solution :
\[E=\frac{12375}{5000}=2.475\,eV\,\approx 4\times {{10}^{-19}}J\] So the minimum intensity to which the eye can respond \[{{I}_{Eye}}=\] (Photon flux) ´ (Energy of a photon) Þ \[{{I}_{Eye}}=(5\times {{10}^{4}})\times \,(4\times {{10}^{-19}})\tilde{}2\times {{10}^{-14}}\,(W/{{m}^{2}})\] Now as lesser the intensity required by a detector for detection, more sensitive it will be \[\frac{{{S}_{Eye}}}{{{S}_{Ear}}}=\frac{{{I}_{Ear}}}{{{I}_{Eye}}}\]\[=\frac{{{10}^{-13}}}{2\times {{10}^{-14}}}=5\] i.e. as intensity (power) detector, the eye is five times more sensitive than ear.
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