- ✓$5$
- B$10$
- C$106$
- D$1539$
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.
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| $\lambda(\mu \mathrm{m})$ | $V_0$ (Volt) |
| $0.3$ | $2.0$ |
| $0.4$ | $1.0$ |
| $0.5$ | $0.4$ |
Given that $c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ and $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant (in units of $\mathrm{J}$ s) found from such an experiment is


