A 1 mW He-Ne laser beam i.e., as in problem (3), has a wavelength lambda = 632.8
ID: 1616840 • Letter: A
Question
A 1 mW He-Ne laser beam i.e., as in problem (3), has a wavelength lambda = 632.8 nm. The amplitude of the electric field for this laser beam is E_0 = 978 V/m. As discussed in class lectures, write the complete mathematical representations for the field E vector (x, t), and the magnetic field, B vector (x, t). These solutions should be written in the form shown below. Using the information provided in class, calculate the values for the following quantities: 1) the magnitude of the magnetic field, B_0; 2) the wave vector magnitude k; and 3) the angular frequency, omega. E vector(x, t) = E vector_0cos(kx + omega t) B vector (x, t) = B vector_0(kx + omega t) The He-Ne laser beam is moving in the x-direction. a. magnitude of B_0 = 3 mT, the wave vector k = 12 times 10^3 rad/m; and omega = 2 times 10^10 rad/sec. b. magnitude of B_0 = 6.8 mu T, the wave vector k = 1 times 10^6 rad/m; and omega = 5 times 10^12 rad/sec. c. magnitude of B_0 = 4 nT, the wave vector k = 22 times 10^2 rad/m; and omega = 4.3 times 10^6 rad/sec. d. magnitude of B_0 = 3.6 mu T, the wave vector k = 10 times 10^6 rad/m; and omega = 3 times 10^15 rad/sec.Explanation / Answer
Eo = 978 V/m
Eo = Boc
978 = Bo*3*10^8
Bo = 3.26*10^-6 T = 3.26 uT
lamda = 632.8 nm
k = 2pi/lamda
k =2*3.14/(632.8*10^-9)
k = 10*10^6 rad/s
w/k = c
w = 10*10^6*3*10^8
w = 3*10^15 rad/s
correct option is (d)
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