Question

A light detector (your eye) has an area of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$and absorbs 80% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.00 m from the source. If the detector absorbs photons at the rate of exactly$\mathbf{4}\mathbf{.}\mathbf{000}\mathbf{}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$at what power does the emitter emit light?

Short answer

The emitter emits power ${\mathrm{1.1\times 10}}^{-10}\mathrm{W}$.

Step-by-step solution

Describe the expression of energy of the photon

The energy Eof a photon of wavelength $\lambda$ is given by,

$E=\frac{hc}{\lambda }$ ……. (1)

Here, h is Planck’s constant, and c is the speed of light.

Determine the power that emitter emit

Let $R_{emit}$be the photon emission rate, then the absorption rate ${\mathrm{R}}_{\mathrm{abs}}=4.00{\mathrm{s}}^{-1}$in terms of the area of the eye ${\mathrm{A}}_{\mathrm{abs}}=2.00\times {10}^{-6}{\mathrm{m}}^2$

$R_{\mathrm{abs}}=\frac{P_{\mathrm{abs}}}{E_p}$

Since the detector absorbs only of the incident light, then the absorbed power is,

$\begin{array}{c}{P}_{\mathrm{abs}}=\left(\frac{80}{100}\right)I{A}_{\mathrm{abs}}\\ R_{\mathrm{abs}}{E}_p=\left(0.8\right)\left(\frac{P_{\mathrm{emit}}}{A_{\mathrm{emit}}}\right)A_{\mathrm{abs}}\\ P_{\mathrm{emit}}=\frac{R_{\mathrm{abs}}{E}_p}{0.8}\left(\frac{4\pi r^2}{A_{\mathrm{abs}}}\right)\end{array}$....(2)

Substitute the below values in eq. 1

$$\begin{gathered} \mathrm{\lambda }=500\mathrm{nm}. \\ \mathrm{h}=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{p}}=\frac{\left({\displaystyle 6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{500\mathrm{nm}} \\ =\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(500\mathrm{nm}\right)\left({\displaystyle \frac{1\mathrm{m}}{{10}^9\mathrm{nm}}}\right)} \\ =3.93\times {10}^{-19}\mathrm{J}. \end{gathered}$$

Substitute the below values in eq. 2.

$$\begin{gathered} {\mathrm{R}}_{\mathrm{abs}}=4.00{\mathrm{s}}^{-1} \\ {\mathrm{E}}_{\mathrm{p}}=3.93\times {10}^{-19}\mathrm{J}. \\ {\mathrm{A}}_{\mathrm{abs}}=2.00\times {10}^{-6}{\mathrm{m}}^2 \\ \mathrm{r}=3\mathrm{m} \end{gathered}$$

$$\begin{gathered} P_{\mathrm{emit}}=\frac{\left({\displaystyle 4.00{\mathrm{s}}^{-1}}\right){\displaystyle \left(3.93\times {10}^{-19}\mathrm{J}\right)}}{0.8}\left(\frac{4\mathrm{\pi }{\left(3\mathrm{m}\right)}^2}{{\displaystyle 2.00\times {10}^{-6}{\mathrm{m}}^2}}\right) \\ =1.1\times {10}^{-10}\mathrm{W} \end{gathered}$$

Therefore, the emitter emits power ${\mathrm{1.1\times 10}}^{-10}\mathrm{W}$.