Remote Sensing Aloysius Roberts

Homework 1, Problems Chapter 2
MWIR radiation covers the 3-5um portion of the EM spectrum. What energy range does this correspond to in eV?
E=h c/ℷ , where h = 4.136×10^-15 eV and c= 3 x 10^8 m/s

Energy range: 0.2482 – 0.4136 eV.
Show that 1um corresponds to roughly 1 eV.

X band radar has a wavelength and frequency that ranges from 2.4 to 3.8 centimeters and 12.5 to 8 GHz.
Ground state energy for a He+ ion in eV? Z = 2, n = 1

E=Z^2 E_1/n^2 where: E1= -13.58 eV.

Calculate E, f, and ℷ for n= 4 to n = 2 transition in a H+ atom.
Energy in eV = -0.85 eV at n = 4. E= -3.39 eV at n = 2.
∆E = n=4 – n=2 = 2.54 eV
Energy emitted in eV for n=4 to n=2 transition equals to 2.54 eV.
ℷ = h*c / ∆E = (〖1.24×10〗^(-6) eVm)/2.54eV=488 nm
F = c / ℷ = 6.14×10^14 Hz
Radiance plot in dependence of wavelength from 0.1um to 20um at different Temperatures Using matlab:

L(ℷ)=(2hc^2)/ℷ^5 *1/(e^(hc/ℷkt)-1)

T=1000:1000:6000;% Kelvin
lambda=[0.1:0.1:20].*10^(-6);% wavelength in micrometers
c=3*10^8;% speed of light in m/s
h=6.626*10^(-34);% J*s
k=1.38*10^(-23);% J/K

for i=1:length(T)
for j=1:length(lambda)
Front_term(j)=(2*h*c^2)/((lambda(j))^5);
Bottom_term(j,i)=(exp((h*c)/(lambda(j)*k*T(i))))-1;
L(j,i)=Front_term(j)*1/Bottom_term(j,i);
end
end

plot(lambda.*10^(-6),L)
legend(‘T=1000′,’T=2000′,’T=3000′,’T=4000′,’T=5000′,’T=6000’)
xlabel(‘Wavelength (um)’)
ylabel(‘Radiance’)
title(‘Problem 6’)

Peak wavelength for radiation at T= 297K, 1000k, and 5800K.
ℷ=a/T where a = 2.898×10^-3(mk)

λ1= 9.7576 um = 9757.6 nm
λ2= 2.898 um = 2898 nm
λ3= 0.49966 um = 499.7nm

Radiated Power=R= σ ℇ T^4

n_1 sinθ_1=n_2 sinθ_2

Since n1 is equal to 1, n2=1.33, and ϴ1 is 30 degreees:
Sinϴ2=sin30/1.33
sinϴ2=0.3759
ϴ2=〖Sin〗^(-1) 0.3759=22.1 degrees
V=C/n2=3*〖10〗^8/1.33=2.255*〖10〗^8
The reflected angle is equal to the angle of incidence: 30 degrees.

R=((n_1-n_2)/(n_1+n_2 ))^2, where n1= 1 and n2 = 1.5

R=(-0.5/2.5)^2=0.04