a) Show formally that:
$$\Delta \omega \approx \frac{area}{R^2}$$by integrating a cone like this:
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from IPython.display import Image
Image('figures/full_triangle.png')
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starting with the exact formula:
$$\Delta \omega = \int d \omega = \int \int \,\sin \theta\, d \theta\, d\phi\ \ \ \ \mathbf{eq: exact}$$with $\theta$ going from 0 to $\theta$ for the small angle case where
$$\sin \theta \approx \theta$$and
$$\cos \theta \approx 1 $$Show that in this case the integral works out to:
$$\Delta \omega = \frac{\pi (r)^2}{R^2} \ \ \ \ \ \mathbf{eq:approx}$$and that therefore if $L=L^*$ everywhere the flux integral
$$ E = 2 \pi \int_0^{\theta} L \cos \theta \sin \theta \, d\theta $$becomes
$$E = L \, \Delta \omega$$Answer: Because we're assuming that $\sin \theta \approx \theta$, and $\cos \theta \approx 1$ we can say:
$R \approx l$ (since $l = R \cos \theta$) so that $\sin \theta = r/R \approx r/l = \theta$.
That means we can write the integral as:
$$\Delta \omega = \int d \omega = \int_0^{2\pi} \int_0^\theta \,\sin \theta^\prime\, d \theta^\prime\, d\phi$$$$\Delta \omega = 2 \pi \int_0^\theta \,\sin \theta^\prime\, d \theta^\prime = 2 \pi \int_0^{r/l} \, \theta^\prime\, d \theta^\prime = 2 \pi \left. \frac{\theta^2}{2} \right |_0^{r/l} = \frac{\pi r^2}{l^2} $$b) Suppose $\theta$ = 5 degrees above. What is the difference between the value of $\Delta \omega$ found by (eq:exact) and (eq:approx)?
Answer: The exact answer:
$$\Delta \omega = 2\pi \int_0^\theta \sin \theta^\prime d\theta^\prime = \left [ -2\pi \cos \theta^\prime \right ]_0^\theta = 2 \pi \left [ 1 - \cos \theta \right ]$$
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import numpy as np
theta = 5*np.pi/180.
r_over_l = np.arctan(theta)
print('r/l = {:7.4f} rads'.format(r_over_l))
approx=np.pi*r_over_l**2.
print('approx omega: {:7.4f} sr'.format(approx))
exact = 2*np.pi*(1 - np.cos(theta))
print('exact omega: {:7.4f} sr'.format(exact))
print('percentage difference={:5.2f}%'.format((exact -approx)/exact*100.))
c) According to the Modis specs the instrument:
Flies at an altitude of 705 km
Makes a cross-track scan in about 0.05 seconds
measures wavelengths with a wavelength range of $\Delta \lambda$ = 11.28 - 10.78 = 0.5 $\mu m$ for channel 31
and according to this link has a detector with the largest detector area = $540 \times 540\ \mu m^2$
the pixel size directly underneath the satellite is 1 $km^2$
Since there are about 1000 pixels in an across-track scan, let's assume that the dwell time on any one pixel is $0.05/1000$ = $5 \times 10^{-5}$ seconds.
Suppose we image a pixel that is emitting a radiance of $L_\lambda = 10\ W\,m^{-2}\,\mu m^{-1}\,sr^{-1}$. Given all of this information, find the energy, in Joules, that the detector receives from that pixel. (Hint: you enough information to calculate detector area, $\Delta \omega$, $\Delta t$, $\Delta \lambda$ and $\Delta area$, and $L_\lambda = Joules /(area \times wavelength\ range \times \Delta t \times \Delta \omega)$)
Answer: step 1: find the solid angle: $\Delta \omega$ = area/$R^2$ = $1\ km^2/(705 km)^2$
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from scipy.constants import c,h
omega = 1/705**2.
print('omega = {:7.4g} sr'.format(omega))
delt=5.e-5 #seconds
del_lambda=0.5 #microns
del_area = 540e-6**2. #m^2
L = 10 #W/m^2/micron/sr
energy = L*del_lambda*delt*del_area*omega
print('energy = {:8.4e} J'.format(energy))
freq = c/11.e-6
photon_energy = h*freq
photon_count = energy/photon_energy
print('this is equal to about {:6d} photons'.format(int(photon_count)))
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