Chapter 3, Problem 3.18P

To determine

To determine the colour indices $U-B$ and $B-V$ for the sun.

Answer to Problem 3.18P

The actual value of ${\left(U-B\right)}_{sun}$ is $0.195$ which is different than the calculated value and the actual value of the sun ${\left(B-V\right)}_{sun}=0.65$ but the calculated value is approximately near the actual value.

Explanation of Solution

Write the expression $U-B$ colour indices.

    $U-B=-2.5{\log }_{10}\left(\frac{B_{365}\Delta {\lambda }_U}{B_{365}\Delta {\lambda }_B}\right)+C_{U-B}$        (1)

Here, ${\lambda }_U$ is wavelength of ultraviolet rays, $B$ is filter and $C_{U-B}$ is colour constant.

Write the expression $B-V$ colour indices.

    $B-V=-2.5{\log }_{10}\left(\frac{B_{440}\Delta {\lambda }_B}{B_{550}\Delta {\lambda }_V}\right)+C_{B-V}$        (2)

Here, $C_{B-V}$ is colour constant.

Write the expression for Planck’s function.

    $B_{\lambda }\left(T\right)=\frac{2h{c}^2}{{\lambda }^5\left(e^{\frac{hc}{\lambda kT}}-1\right)}$        (3)

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

Write the expression for $U$ filter using equation (3).

    $B_{\lambda U}\left(T\right)=\frac{2h{c}^2}{{\lambda }_U{}^5\left(e^{\frac{hc}{{\lambda }_{U}kT}}-1\right)}$        (4)

Write the expression for $B$ filter using equation (3).

    $B_{\lambda B}\left(T\right)=\frac{2h{c}^2}{{\lambda }_B{}^5\left(e^{\frac{hc}{{\lambda }_{B}kT}}-1\right)}$        (5)

Write the expression for $V$ filter using equation (3).

    $B_{\lambda V}\left(T\right)=\frac{2h{c}^2}{{\lambda }_V{}^5\left(e^{\frac{hc}{{\lambda }_{V}kT}}-1\right)}$        (6)

Divide equation (6) by (5) and solve with equation (2).

    $\begin{array}{l}\frac{B_{\lambda V}\left(T\right)}{B_{\lambda B}\left(T\right)}=\frac{\frac{2h{c}^2}{{\lambda }_V{}^5\left(e^{\frac{hc}{{\lambda }_{V}kT}}-1\right)}}{\frac{2h{c}^2}{{\lambda }_B{}^5\left(e^{\frac{hc}{{\lambda }_{B}kT}}-1\right)}}\\ B-V=-2.5{\log }_{10}\left(\frac{{\lambda }_V}{{\lambda }_B}\right)\left(\frac{e^{\frac{hc}{{\lambda }_{V}kT}}-1}{e^{\frac{hc}{{\lambda }_{B}kT}}-1}\right)\left(\frac{\Delta {\lambda }_B}{\Delta {\lambda }_V}\right)+C_{B-V}\end{array}$        (7)

Divide equation (5) by (4) and solve with equation (2).

    $\begin{array}{l}\frac{B_{\lambda B}\left(T\right)}{B_{\lambda U}\left(T\right)}=\frac{\frac{2h{c}^2}{{\lambda }_B{}^5\left(e^{\frac{hc}{{\lambda }_{B}kT}}-1\right)}}{\frac{2h{c}^2}{{\lambda }_U{}^5\left(e^{\frac{hc}{{\lambda }_{U}kT}}-1\right)}}\\ U-B=-2.5{\log }_{10}{\left(\frac{{\lambda }_B}{{\lambda }_U}\right)}^5\left(\frac{e^{\frac{hc}{{\lambda }_{B}kT}}-1}{e^{\frac{hc}{{\lambda }_{U}kT}}-1}\right)\left(\frac{\Delta {\lambda }_U}{\Delta {\lambda }_B}\right)+C_{U-B}\end{array}$        (8)

Conclusion:

Substitute $440\text{nm}$ for ${\lambda }_B$ and $365\text{nm}$ for ${\lambda }_U$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $1.38\times {10}^{-23}\text{J}/\text{k}$ for $k$, $5777\text{K}$ for $T$, $68\text{nm}$ for $\Delta {\lambda }_U$ and $98\text{nm}$ for $\Delta {\lambda }_B$ equation (7).

    $\begin{array}{c}U-B=-2.5{\log }_{10}{\left(\frac{365\text{nm}}{440\text{nm}}\right)}^5\left(\frac{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(440\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(365\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}\right)\left(\frac{68\text{nm}}{98\text{nm}}\right)-0.87\\ =-2.5{\log }_{10}{\left(\frac{365\text{nm}}{440\text{nm}}\right)}^5\left(\frac{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(440\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(365\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}\right)\left(0.693877\right)-0.87\\ =-2.5{\log }_{10}\left(0.39282\right)\left(\frac{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(440\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(365\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}\right)\left(0.693877\right)-0.87\\ =-0.22\end{array}$

Substitute $440\text{nm}$ for ${\lambda }_B$ and $550\text{nm}$ for ${\lambda }_V$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $1.38\times {10}^{-23}\text{J}/\text{k}$ for $k$, $5777\text{K}$ for $T$, $89\text{nm}$ for $\Delta {\lambda }_V$ and $98\text{nm}$ for $\Delta {\lambda }_B$ equation (8).

    $\begin{array}{c}U-B=-2.5{\log }_{10}{\left(\frac{550\text{nm}}{440\text{nm}}\right)}^5\left(\frac{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(500\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(440\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}\right)\left(\frac{98\text{nm}}{89\text{nm}}\right)+0.65\\ =-2.5{\log }_{10}{\left(\frac{550\text{nm}}{440\text{nm}}\right)}^5\left(\left(\frac{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(500\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}{e^{\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(440\text{nm}\right)\left(1.38\times {10}^{-23}\text{J}/\text{k}\right)\left(5777\text{K}\right)}}-1}\right)\right)\left(1.10112\right)-+0.65\\ =0.57\end{array}$

Thus, the actual value of ${\left(U-B\right)}_{sun}$ is $0.195$ which is different than the calculated value and the actual value of the sun ${\left(B-V\right)}_{sun}=0.65$ but the calculated value is approximately near the actual value.