Can someone please help me understand how we arrived at the answers? I\'m pretty

Question

Can someone please help me understand how we arrived at the answers? I'm pretty lost :/

A laser beam falls on an area A=0.00250 m^2. The equation of the electric field corresponding to the sinusoidal (plane) waves in the beam is represented by: E = 4.650 Times 10^5 (V/m) sin[omega (rad/s) t - 9.926 Times 10^6 (rad/m) x]. Calculate: (a) the Emax; (b) the Emax; (c) the constant k and (d) its wavelength lambda; (e) the frequency f and (f) its corresponding omega; (g) the average energy density uave and (h) equivalent radiation pressure p ; (i) the intensity I; (j) the power P falling on the given area; (k) the total energy delivered in 0.200 seconds; (l) the force F on the same area; (m) tell the direction of the wave and (n) if it is visible by humans. Fill the results in the Table below using four significant figures. Answers: 4.650 Times 10^5; 1.550 Times 10^3; 9.926 Times 10^6 6.330 Times 10^7; 4.739 Times 10^14; 2.978 Times 10^15 0.95721(h)?; 2.872; 2.500 Times 10^3; 5.000 Times 10^-4; 2.393 Times 10^-3

Explanation / Answer

relation between Electric field and magnetic field is E = Bc

so

part A: Emax = 4.65 e 5 V/m

part B: Bmax = E/c = 4.65 e 5 /3e8 = 1.55e -3 Tesla

part C: Cosnstant K upon comparing k = 9.926 e 6 rad/m

part D : Wavelength = 2pi/K = 2*3.14/(9.926e 6) = 6.33e -7 m

part E : use V = Lf ---> frequency f = V/L = 3e8/(6.33 e-7) = 4.739 e14 Hz

Part F: W = 2pif = 2*3.14 * 4.739e 14 = 2.978 e 15 rad/s

part G: Eenrgy Densty u = 0.5 eo E^2 = 0.5* 8.85 e -12 * 4.65 e 5 * 4.65 e 5 = 0.9567 J/m^3

part I : Intensity I = power/area = Emax^2/2uoC = (4.65 e 5*4.65 e 5)/(2*4*3.14 e-7 * 3e8) = 2.86 e 8

part H:Radiation pressure = 2I/C = 2* 2.86 e 8/3e8 = 1.91 Pa

part J : Power P = I*A = 2.86 e 8 * 0.0025 = 7.15e+5 Watts

part K : Eenrgy E = Power * time = 7.15 e5 * 0.2 = 1.43e+5 Joules

part L: Force F = Pressure * area = 1.91* 0.0025 = 4.773 e-3 N

Part M: direction = - x axis

part N : Wavelength L = V/f = 3e8 /4.739e 14 = 633 nm . yes it visible Light

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