Chapter 3, Problem 63P

(a)

To determine

The number of photons that enter the telescope.

Answer to Problem 63P

The number of photons detected by the detector is $0.015\text{photons}/\text{s}$.

Explanation of Solution

Write the expression for the power of the incident photons per second on the detector.

  $P=I\cdot A$

Here, $I$ is the intensity of the photon, $A$ is the area of cross-section and $P$ is the incident power.

Write the expression for the energy per second detected by the detector of telescope.

  $E=P\cdot t$

Substitute $I\cdot A$ for $P$ in above expression.

  $E=\left(I\cdot A\right)\cdot t$        (I)

Here, $E$ is the energy incident per second and $t$ is the time.

Write the expression for the energy of the incident photon of particular wavelength.

  $E^{\prime }=\frac{hc}{\lambda }$        (II)

Here, $E^{\prime }$ is the energy of one photon, $h$ is the Planck’s constant, $c$ is the speed of the light and $\lambda$ is the incident wavelength.

Write the expression for the number of photons incident on the detector.

  $n=\frac{E}{E^{\prime }}$        (III)

Here, $n$ is the number of photons.

Conclusion:

Substitute $2\times {10}^{-20}\text{W}/{\text{m}}^{\text{2}}$ for $I$, $0.30{\text{m}}^{\text{2}}$ for $A$ and $1\text{s}$ for $t$ in equation (I).

  $\begin{array}{c}E=\left(2\times {10}^{-20}\text{W}/{\text{m}}^{\text{2}}\right)\cdot \left(0.30{\text{m}}^{\text{2}}\right)\cdot 1\text{s}\\ =6\times {10}^{-21}\text{J}/\text{s}\end{array}$

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $486\times {10}^{-9}\text{m}$ for $\lambda$ in equation (II).

  $\begin{array}{c}{E}^{\prime }=\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{486\times {10}^{-9}\text{m}}\\ =4.09\times {10}^{-19}\text{J}\end{array}$

Substitute $6\times {10}^{-21}\text{J}/\text{s}$ for $E$ and $4.09\times {10}^{-19}\text{J}$ for $E^{\prime }$ in equation (III).

  $\begin{array}{c}n=\frac{6\times {10}^{-21}\text{J}/\text{s}}{4.09\times {10}^{-19}\text{J}}\\ =0.015\text{photons}/\text{s}\end{array}$

Thus, the number of photons detected by the detector is $0.015\text{photons}/\text{s}$.

(b)

To determine

The number of photons reaching the retina at  the $6^{th}$ magnitude of the star.

Answer to Problem 63P

The number of photons detected by the detector is $6.45\times {10}^3\text{photons}/\text{s}$.

Explanation of Solution

 Write the expression for the intensity of the $6^{th}$ magnitude object.

  $I=I_0\cdot {\left({100}^{1/5}\right)}^{24}$        (IV)

Here, $I^{\prime }$ is the intensity of the $6^{th}$ magnitude of the object and $I_0$ is the intensity of ${30}^{th}$ magnitude.

Conclusion:

Substitute $2\times {10}^{-20}\text{W}/{\text{m}}^{\text{2}}$ for $I$ in equation (IV).

  $\begin{array}{c}{I}^{\prime }=\left(2\times {10}^{-20}\text{W}/{\text{m}}^{\text{2}}\right)\cdot {\left(2.5119\right)}^{24}\\ =\left(2\times {10}^{-20}\text{W}/{\text{m}}^{\text{2}}\right)\cdot \left(3.98\times {10}^9\right)\\ =7.96\times {10}^{-11}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Substitute $7.96\times {10}^{-11}\text{W}/{\text{m}}^{\text{2}}$ for $I$, $6.5\text{mm}$ for $d$ and $1\text{s}$ for $t$ in equation (I).

  $\begin{array}{c}E=\left(7.96\times {10}^{-11}\text{W}/{\text{m}}^{\text{2}}\right)\cdot \left(\pi {\left(\frac{6.5\text{mm}}{2}\right)}^2\right)\cdot 1\text{s}\\ =2.64\times {10}^{-15}\text{J}/\text{s}\end{array}$

Substitute $2.64\times {10}^{-15}\text{J}/\text{s}$ for $E$ and $4.09\times {10}^{-19}\text{J}$ for $E^{\prime }$ in equation (III).

  $\begin{array}{c}n=\frac{2.64\times {10}^{-15}\text{J}/\text{s}}{4.09\times {10}^{-19}\text{J}}\\ =6.45\times {10}^3\text{photons}/\text{s}\end{array}$

Thus, the number of photons detected by the detector is $6.45\times {10}^3\text{photons}/\text{s}$.