Chapter 5, Problem 5P

(1)

To determine

The diameter of pupil of human eye needed to see the same resolution in the infrared at wavelength of $12\mu \text{m}$ as he can in the visible at $500\text{nm}$.

Answer to Problem 5P

The diameter of pupil of human eye needed to see the same resolution in the infrared at wavelength of $12\mu \text{m}$ as he can in the visible at $500\text{nm}$ is $\underset{\_}{12\text{cm}}$.

Explanation of Solution

The diameter of pupil of human eye is $5\text{mm}$, and the wavelength in the visible range is given as $500\text{nm}$.

Write the expression for angle of separation between two objects according to the criterion for resolution.

  $\alpha =\frac{0.02\lambda }{D}$        (I)

Here, $\alpha$ is the angle of separation, $\lambda$ is the wavelength, and $D$ is the diameter.

Since the person need to see the same resolution in the infrared as in visible the angle of separation in both region of light can be equated.

  ${\alpha }_{\text{visible}}={\alpha }_{\text{infrared}}$        (II)

Using equation (I), equation (II) can be modified as,

  $\frac{0.02{\lambda }_{\text{visible}}}{D_{\text{visible}}}=\frac{0.02{\lambda }_{\text{infrared}}}{D_{\text{infrared}}}$        (III)

Solve equation (III) for $D_{\text{infrared}}$.

  $D_{\text{infrared}}=\frac{{\lambda }_{\text{infrared}}{D}_{\text{visible}}}{{\lambda }_{\text{visible}}}$        (IV)

Conclusion:

Substitute, $12\mu \text{m}$ for ${\lambda }_{\text{infrared}}$, $5\text{mm}$ for $D_{\text{visible}}$, $500\text{nm}$ for ${\lambda }_{\text{visible}}$, in equation (IV) to find $D_{\text{infrared}}$.

  $\begin{array}{c}{D}_{\text{infrared}}=\frac{\left(12\mu \text{m}\right)\left(5\text{mm}\right)}{500\text{nm}}\\ =\frac{\left(12\mu \text{m}\times \frac{1\text{cm}}{1\times {10}^4\text{cm}}\right)\left(5\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\right)}{500\text{nm}\times \frac{1\text{cm}}{1\times {10}^7\text{nm}}}\\ =12\text{cm}\end{array}$

Therefore, the diameter of pupil of human eye needed to see the same resolution in the infrared at wavelength of $12\mu \text{m}$ as he can in the visible at $500\text{nm}$ is $\underset{\_}{12\text{cm}}$.

(2)

To determine

The diameter of pupil of human eye needed to see the same resolution in the radio at wavelength of $10\text{cm}$ as he can in the visible at $500\text{nm}$.

Answer to Problem 5P

The diameter of pupil of human eye needed to see the same resolution in the radio at wavelength of $10\text{cm}$ as he can in the visible at $500\text{nm}$ is $\underset{\_}{1\text{km}}$.

Explanation of Solution

The diameter of pupil of human eye is $5\text{mm}$, and the wavelength in the visible range is given as $500\text{nm}$.

Since the person need to see the same resolution in the radio as in visible the angle of separation in both region of light can be equated.

  ${\alpha }_{\text{visible}}={\alpha }_{\text{radio}}$        (V)

Similar to equation (III) it can be written for the visible and radio wave lights as,

  $\frac{0.02{\lambda }_{\text{visible}}}{D_{\text{visible}}}=\frac{0.02{\lambda }_{\text{radio}}}{D_{\text{radio}}}$        (VI)

Solve equation (VI) for $D_{\text{radio}}$.

  $D_{\text{radio}}=\frac{{\lambda }_{\text{radio}}{D}_{\text{visible}}}{{\lambda }_{\text{visible}}}$        (VII)

Conclusion:

Substitute, $10\text{cm}$ for ${\lambda }_{\text{radio}}$, $5\text{mm}$ for $D_{\text{visible}}$, $500\text{nm}$ for ${\lambda }_{\text{visible}}$, in equation (VII) to find $D_{\text{radio}}$.

  $\begin{array}{c}{D}_{\text{radio}}=\frac{\left(10\text{cm}\right)\left(5\text{mm}\right)}{500\text{nm}}\\ =\frac{\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(5\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{500\text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{nm}}}\\ =1000\text{m}\times \frac{1\text{km}}{1000\text{m}}\\ =1\text{km}\end{array}$

Therefore, the diameter of pupil of human eye needed to see the same resolution in the radio at wavelength of $10\text{cm}$ as he can in the visible at $500\text{nm}$ is $\underset{\_}{1\text{km}}$.