Chapter 27, Problem 71A

(a)

To determine

To Find: The power of light entering into person’s eye.

Answer to Problem 71A

  $5.77\times {10}^{-14}\text{W}$

Explanation of Solution

Given data:

Intensity of visible light $=1.5\times {10}^{-11}\frac{\text{W}}{{\text{m}}^{\text{2}}}$

Diameter of pupil, $d=70\text{mm = 70}\times {\text{10}}^{-3}\text{m}$

Formula used:

Power can be written in terms of intensity and area:

  $P=I\times A$

Here, I is the intensity of light and A is the area.

Calculation:

  $\begin{array}{l}r=\frac{d}{2}\\ r=\frac{70}{2}\\ r=35\text{mm = 35}\times {\text{10}}^{-3}\text{m}\end{array}$

  $\begin{array}{l}$ A=\pi {(\text{r)}}^2$$ \text{ =}\pi {(35\times {10}^{-3})}^2{\text{m}}^{\text{2}}$$ P=1.5\times {10}^{-11}\times {(35\times {10}^{-3})}^2\times \pi $\text{ = }5.77\times {10}^{-14}\text{W}\end{array}$

Conclusion:

Thus, the power of light that enters the eye is $5.77\times {10}^{-14}\text{W}$ .

(b)

To determine

To Calculate: The number of photons entering into person’s eye if wavelength of entering light is 550 nm and diameter of pupils is 550nm in given figure.

Answer to Problem 71A

  $1.6\times {10}^5{\text{s}}^{-1}$

Explanation of Solution

Given data:

Intensity of visible light $=1.5\times {10}^{-11}\frac{\text{W}}{{\text{m}}^{\text{2}}}$

Wavelength of light, $\lambda =550\text{nm}=550\times {10}^{-9}\text{m}$ .

The power of light that enters the eye, $P=5.77\times {10}^{-14}\text{W}$ .

Formula used:

Energy of each photon, $(E)=\frac{hc}{\lambda }$

Here, h is the Planck’s constant, c is the speed of light and $\lambda$ is the wavelength.

Number of photons per second entering into eye $=\frac{\text{power}}{\text{Energy of photon}}$

Calculation:

Energy of each photon,

  $\begin{array}{l}(E)=\frac{hc}{\lambda }\\ E=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{550\times {10}^{-9}}=3.6\times {10}^{-19}\text{J}\end{array}$

Number of photons per second entering the eye:

  $\begin{array}{l}=\frac{\text{Power}}{\text{Energy}}\\ =\frac{5.77\times {10}^{-14}\text{J/s}}{3.6\times {10}^{-19}\text{J}}=1.6\times {10}^5{\text{s}}^{-1}\end{array}$

Conclusion:

Therefore, number of photons per second entering into the person’s eye is $1.6\times {10}^5{\text{s}}^{-1}$ .