# Field trip

NEW YORK CITY COLLEGE OF TECHNOLOGY
ELECTRICAL AND TELECOMMUNICATIONS ENGINEERING TECHNOLOGY DEPARTMENT
EET 3132: Remote Sensor home work
Student: ALOYSIUS ROBERTS
May 16, 2016

Problem 1) At 3µm, E=hc/lambda =(4.14*〖10〗^(-15eVs)*3*10^8(m/s))/(3*10^-6m) =0.413eV
At 5µm, E = (4.14*〖10〗^(-15eVs)*3*10^8(m/s))/(5*10^-6m) 0.24eV
3µm – 5µm
0.413eV – 0.24eV
Problem 2)Δ
At 1µm, E = hf = hc/lambda = (4.14*〖10〗^(-15eVs)*3*10^8(m/s))/(1*10^-6m) =1.24eV
Problem 3)
Frequency Wavelength

Problem 4)
E = 1/2((Ze^2)/4π€oh)m/n^2 = Z^2 E_1/n^2 = 2^2((-13.58)/1^2 ) = -54.32eV
Problem 5)
E_4 = Z_4^2 = E_1/(n_4^2 ) = (-13.58eV)/4^2 = -0.84875eV
E_2= = Z_2^2 E_1/(N_2^2 ) =(-13.58eV)/2^2 =-3.395Ev
 = hc/ΔE = (1.24*10^-6eVm)/(-0.8485eV-(-3.395eV)) = 486nm
f = c/lambda = (3*10^8(m/s))/(486*10^-9m) = 6.16*10^14Hz
Problem 7)
At T = 297K, (max) = a/T = (2.898*10^-3(mk))/297k = 9.76*10^-6m
At T = 1000K , (max) =a/T = (2.898*10^-3(mk))/1000k = 2.898*10^-6m
At T = 5800K, (max) = a/T = (2.898*10^-3(mk))/5800k = 0.499*10^-6m
Problem 8)
Stefan-Bolzmann law: R = ^4(Watts/m^2 )
R = 5.67*10^-8(w/(m^2k^4))(1)(297k^4)2m^2 = 882.35 watts
Problem 9)
x=0:0.1:100;
y=x.^5/(exp(x)-1);
y_max=max(y)
c_index=find(y==y_max);
x_max=x(c_index);
plot(x,y)
axis([0 20 0 25]);

>> x=0:1:100;
>> y=x.^5/(exp(x)-1);
>> plot(x,y)
>> x=0:0.1:100;
>> y=x.^5./(exp(x)-1);
>> y_max=max(y)

y_max =

21.1989

>> c_index=find(y==y_max);
>> x_max=x(c_index);
plot (x,y)

L = (2πhc^2)/(^5) 1/〖e(hc/kT)-1〗^ = 〖2x〗^5/(c^(3h^4 )*e^x*(-1)) 〖KT〗^5
X = hc/KT
 = hc/XKT = hc/(XK/T)
(max)= a/(T ); a = hc/KT = (6.626*〖10〗^(-34(j.s)(3*〖10〗^8 )m)/s))/(5(1.38*5^10-23(J/K)) = 2.880869mK

Problem 10)
n_1 = 1; n_(2=1.33)
n_1 〖sinθ〗_1 = n_2 〖sin〗_(2;)1*sin〖30〗^degrees = 1.33〖sin〗_2;1.33sinθ2=0.5
Θ2 = 〖sin〗^(-1)(0.3759)=22.07degrees

Problem 11)
R = (〖(n1-n2)/(n1+n2))〗^2 =(1-1.5)/(1+1.5) =(〖-0.2)〗^2 = 0.04