Chapter 9, Problem 9.1P

To determine

The energy of blackbody photons inside the eye, and compare it with the visible energy inside the eye while looking at a 100-W light bulb.

Answer to Problem 9.1P

The energy of blackbody photons inside the eye is $9.87\times {10}^{-12}\text{J}$, and the comparison shows energy of bulb is less than energy of pupil’s eyes.

Explanation of Solution

Write the expression for density of radiation of black body per unit volume.

    $a{T}^4=u$        (1)

Here, $u$ is density of radiation of black body, $a$ is absorption constant and $T$ is absolute temperature.

Write the expression for energy density of the photon.

    $E=a{T}^{4}V$        (2)

Here, $V$ is the volume of photons, $a$ is absorption constant and $T$ is absolute temperature.

Write the expression for the flux.

    $F=\frac{1}{4r^2\pi }$        (3)

Here, $r$ is radius.

Write the expression for time required to cross the eye ball.

    $t=\frac{2r_{eye}}{c}$        (4)

Here, $c$ is speed of light and $r_{eye}$ is radius of eye.

Write the expression for energy enter in the eye of person.

    $\begin{array}{c}{E}_{\text{bulb}}=\frac{FA}{t}\\ =\frac{2AL{r}_{eye}}{4c\pi r^2}\end{array}$        (5)

Write the expression of volume of sphere.

    $V=\frac{\pi r_{eye}^{3}4}{3}$        (6)

Here, $r_{eye}$ is radius of eye.

Write the expression for win’s displacement law,

  ${\lambda }_{\text{max}}=\frac{0.002898\text{m}\cdot \text{K}}{T}$        (7)

Here, $T$ is absolute temperature

Conclusion:

The dimension of radius of pupil’s eye in meters is,

    $\begin{array}{c}{r}_{eye}=1.5\text{cm}\\ =0.015\text{m}\end{array}$

Temperature of pupil’s eye is,

    $\begin{array}{c}T=37°\text{C}\\ \text{=310}\text{K}\end{array}$

Substitute $0.015\text{m}$ for $r_{eye}$ in equation (6)

    $\begin{array}{c}V=\frac{\pi {\left(0.015\right)}^{3}4}{3}\\ =\frac{4.279\times {10}^{-5}}{3}\\ =1.4265\times {10}^{-05}\text{}{\text{m}}^{\text{3}}\end{array}$

Substitute $310\text{K}$ for $T$ and $1.4265\times {10}^{-05}$ for $V$ in equation (2)

    $\begin{array}{c}E=\left(310\text{K}\right)\left(7.5657\times {10}^{-16}\text{J}/{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{4}}\right)\left(1.413\times {10}^{-6}{\text{m}}^{\text{3}}\right)\\ =9.87\times {10}^{-12}\text{J}\end{array}$

Thus, the energy of black body is $9.87\times {10}^{-12}\text{J}$

Substitute $0.001{\text{m}}^{\text{2}}$ for $A$, $1\text{m}$ for $r$, $0.015\text{m}$ for $r_{eye}$, $100\text{W}$ for $L$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (4)

    $\begin{array}{c}{E}_{\text{bulb}}=\frac{2\left(0.001{\text{m}}^{\text{2}}\right)\left(100\text{W}\right)\left(0.015\text{m}\right)}{4\left(3\times {10}^8\text{m}/\text{s}\right)\pi {\left(1\right)}^2}\\ =7.96\times {10}^{-13}\text{J}\end{array}$

Hence the comparison shows energy of bulb is less than energy of pupil’s eyes.

Substitute, $310\text{K}$ for $T$ in equation (7),

    $\begin{array}{c}{\lambda }_{\max }=\frac{\left(\frac{1\times {10}^{-3}\text{nm}}{1\text{m}}\right)\left(9.34\times {10}^{-6}\text{}\text{m}\right)}{310\text{K}}\\ =9.34\text{nm}\end{array}$

Thus the wave length calculated in above calculation comes under the infrared radiation, hence we observed darkness at time of close our eyes.