Question

The eye can detect $5\times {10}^4$ photons per square meter per second of green light $(\lambda =5000\mathring{A})$ while the ear can detect sound of intensity ${10}^{-13}{\text{Wm}}^{-2}$. Find the factor by which the eye is more sensitive as a power detector than the ear-

• A. $2$
• B. $3$
• C. $1$
• D. $5$

Answer 1

The correct option is D $5$

Given, $N=5\times {10}^4;\lambda =5000\mathring{A};I_{ear}={10}^{-13}{\text{W m}}^{-2}$

The energy of each photon is,

$E=\frac{hc}{\lambda }=\frac{12400}{5000}=2.48\text{eV}[\because hc=12400\mathring{A}\text{(eV)}]$

$\therefore =3.968\times {10}^{-19}\text{J}$

Photon flux $\left(N\right)$ is given by,

$N=\frac{\text{Intensity}}{\text{Energy of each photon}}=\frac{I}{E}$

$I_{eye}$ = Photon flux $\times$ Energy of a photon

$=(5\times {10}^4)\times (3.968\times {10}^{-19})$

$\approx 2\times {10}^{-14}{\text{Wm}}^{-2}$

Lesser the intensity, more the sensitivity.

$\frac{S_{eye}}{S_{ear}}=\frac{I_{ear}}{I_{eye}}=\frac{{10}^{-13}}{2\times {10}^{-14}}=5$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.