a
(a) $E = \frac{{12375}}{{5000}} = 2.475\,eV\, \approx 4 \times {10^{ - 19}}J$
So the minimum intensity to which the eye can respond
${I_{Eye}} = $ (Photon flux) $×$ (Energy of a photon)
$⇒$ ${I_{Eye}} = (5 \times {10^4}) \times \,(4 \times {10^{ - 19}})\tilde --2 \times {10^{ - 14}}\,(W/{m^2})$
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.