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
$$E = L \, \Delta \omega$$b) Suppose $\theta$ = 5 degrees above. What is the difference between the value of $\Delta \omega$ found by (eq:exact) and (eq:approx)?
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 $L_\lambda = Joules /(area \times wavelength\ range \times \Delta t \times \Delta \omega)$)
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