34. Assignment 3: solutions

WH 4.43, 4.46, 4.47

34.1. Question 4.43

Consider radiation with wavelength λ and zero zenith angle passing through a gas with an absorption coefficient of \(k_\lambda\) = 0.01 \(m^2\,kg^{-1}\). What fraction of the beam is absorbed in passing through a layer containing 1 \(kg\,m^{-2}\) of the gas? What mass of gas would the layer have to contain in order to absorb half the incident radiation?

34.1.1. 4.43 Answer

The equation: use eq. 4.31 for constant absorption coefficient, which can come out of the integral. Since they don’t mention a mixing ratio \(r\), we can assume that it’s the only gas, so \(r\)=1. Also \(\theta = 0\) so \(\sec \theta = 1\). Assume the setup is that the radiation is coming down in a direct beam from above (sunlight), so we can define the mass above z as \(M\ (kg\,m^{-2}\)) where:

\[ M = \int_z^\infty \rho dz \]

Then Beer’s law says:

\[ \tau = k_\lambda \int_z^{\infty} \rho d z = k_\lambda M \]

\[ \frac{I_{\lambda}}{I_{\lambda \infty}} = \exp(-\tau) = \exp \left ( -k_\lambda M \right ) \]

with \(\sec \theta\)=1 and and absorption coefficient \(k_\lambda\) both constant with \(z\)

The numbers:

\[ k_\lambda M = 0.01\ m^2\,kg^{-1} \times 1\ kg\,m^{-2} = 0.01 \]

\[ \frac{I_{\lambda}}{I_{\lambda \infty}} = \exp(-0.01) = 0.37 \]

Column mass needed for 0.5 absorption:

\[\begin{split} 0.5 &= \exp( -0.01 \times M ) \\ \ln(0.5) & = -0.01 \times M \\ M &= -0.69/-0.01 = 69\ kg\,m^{-2} \end{split}\]

import numpy as np
print(f"{np.exp(-0.01)=:.2f}")
print(f"{np.log(0.5)=:.2f}")

np.exp(-0.01)=0.99
np.log(0.5)=-0.69

34.2. Question 4.46

Consider a hypothetical planetary atmosphere comprised entirely of the gas in Exercise 4.43. The atmospheric pressure at the surface of the planet is 1000 hPa , the lapse rate is isothermal, the scale height is 10 km , and the gravitational acceleration is 10 \(m\,s^{-2}\). Estimate the height and pressure of the level of unit normal optical depth.

34.2.1. 4.46 Answer

Again, the mass of the column as \(M = \int \rho dz\)

\[\begin{split} \tau = \int_z^\infty k_\lambda \rho dz = k_\lambda \int_z^\infty \rho dz= k_\lambda M = 1 \\ M = 1/k_\lambda = 1/0.01 = 100\ kg\,m^{-2} \end{split}\]

Hydrostatic equation:

\[\begin{split} \int_{p}^0 dp^\prime &= -\int_0^z \rho g dz^\prime \\ -p &= -g \int_z^\infty \rho dz^\prime = - g M = 10 \times 100 = -1000\ Pa\\ p &= 1000\ Pa \end{split}\]

Height at 1000 Pa given \(p_0 = 10^5\ Pa\) and H = 10 km

\[\begin{split} p &= p_0 \exp(-z/H) \\ 1000 &= 10^5 \exp(-z/H) \\ \ln(10^{-2}) &= -z/H = -z/10\ km \\ -4.6 \times 10 &= - z \\ z &= 46\ km \end{split}\]

np.log(1.e-2)

np.float64(-4.605170185988091)

34.3. Question 4.47

(a) What percentage of the incident monochromatic intensity with wavelength λ and zero zenith angle is absorbed in passing through the layer of the atmosphere extending from an optical depth τ_λ=0.2 to τ_λ=4.0 ? (b) What percentage of the outgoing monochromatic intensity to space with wavelength λ and zero zenith angle is emitted from the layer of the atmosphere extending from an optical depth τ_λ=0.2 to τ_λ=4.0 ? (c) In an isothermal atmosphere, through how many scale heights would the layer in (a) and (b) extend?

Here’s the setup:

Fig. 34.1 Setup for Problem 4.47

34.3.1. 4.47 Answer

a) As with Optical depth II: mean free path, to find the amount absorbed we just need to take the difference between the amount arriving at \(\tau = 0.2\) and the amount that makes it to \(\tau = 4\). So if the fraction absorbed is \(f\), then

\[ f = \frac{I_0 \exp(-0.20 ) - I_0\exp(-4)}{I_0} = .818 - .18 = 0.8 = 80\% \]

np.exp(-0.2), np.exp(-4)

(np.float64(0.8187307530779818), np.float64(0.01831563888873418))

b) Assume that the layer is thick enough so that for the entire layer, emissivity \(\epsilon\) = 1, i.e. the total optical depth \(\tau_T\) is very large. Also assume that we’re again dealing with an isothermal layer. Then the question can be phrased as “what’s the emissivity of the part of the atmosphere between \(\tau = 0.2\) and \(\tau = 4\)?”. If we call that emissivity \(\epsilon_L\), then for an isothermal atmosphere the emitted radiance is going to be \(\epsilon_L B_\lambda\).

First find the emissivity between \(\tau = 0\) and \(\tau = 4\). We know that emissivity = absorptivity, and without reflection absorptivity = 1 - transmissivity. So the emissivity from \(\tau = 0 \rightarrow 4\) is:

\[ \epsilon_{0-4} = 1 - \exp(-4) \]

and the emissivity from \(0-0.2\) is

\[ \epsilon_{0-0.2} = 1 - \exp(-0.2) \]

So the total emitted from \(\tau = 0 \rightarrow 4\) is \(\epsilon_{0-4} B_\lambda\) and the amount emitted from \(\tau = 0\rightarrow0.2\) is \(\epsilon_{0-0.2} B_\lambda\). The fraction f emitted is just the difference between those two amounts divided by the total blackbody emission \(B_\lambda\):

\[ f = \frac{ (1 - \exp(-4)) B_\lambda - (1 - \exp(-0.2))B_\lambda}{B_\lambda} = 0.818 - 0.18 = 0.8 \]

i.e., just another way of saying that 1 - transmissivity = absorptivity = emissivity

c) Assume the density scale height is \(H\), so that

\[ \rho(z) = \rho_0 \exp(-z/H) \]

From part a) the defintion of the optical depth is

\[ \tau = k_\lambda \int_z^{\infty} \rho d z = k_\lambda \int_z^{\infty} \rho_0 \exp(-z/H) d z \]

Integrating this:

\[ \tau = -H k_\lambda \rho_0 \exp(-z/H) \big |_z^\infty = k_\lambda \rho_0 \left ( 0 - (-H \exp(-z/H)) \right ) = k_\lambda \rho_0 H \exp(-z/H) \]

If we say that \(\tau_1 = 4\) occurs at height \(z_1\) and \(\tau_2 = 0.2\) occurs at heigh \(z_2\), then

\[ \frac{\tau_2}{\tau_1} = \frac{0.2}{4} = \frac{k_\lambda \rho_0 H\exp(-z_2/H)}{k_\lambda \rho_0 H \exp(-z_1/H)} = \exp \left ( -\frac{(z_2 - z_1 ) }{H} \right ) \]

Take the log of both sides:

\[ \ln(0.05) = -3 = -\frac{(z_2 - z_1 ) }{H} \]

So the difference between the heights is 3 scale heights.

np.log(0.05)

np.float64(-2.995732273553991)