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The eye can detect $5 ×10^4$ photons per square metre per sec of green light ($\lambda$ $= 5000\ \mathop A\limits^o $) while the ear can detect ${10^{ - 13}}\,(W/{m^2})$. The factor by which the eye is more sensitive as a power detector than the ear is close to
$5$
$10$
$106$
$1539$
Solution
(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.
Similar Questions
Match the column
| $(A)$ Hallwachs $\&$ Lenard | $(P)$ Transformers |
| $(B)$ Franck-Hertz | $(Q)$ Microwave |
| $(C)$ Klystron valve | $(R)$ Quantization of energy levels |
| $(D)$ Nicola Tesla | $(S)$ Photoelectric effect |
