A radio station has two antennas. The antennas are a distance d apart, where d equals half the broadcast wavelength. The antennas are driven in phase with each other. Let the x-axis be the line that runs through the two antennas. The angles are all measured counterclockwise from the +x-direction. (For the following, assume an observer is positioned a distance D far from the midpoint of the antennas, so that  D ≫ d.) (a) In which directions is the strongest signal radiated? 0°, 180° 90°, 270°    0°, 90°, 180°, 270° 45°, 135°, 225°, 315° (b)

Problem 1. In class we will consider the sum of the electric field of two plane waves, u:(z.t) and uz(z.t), both traveling in the positive z direction with slightly different frequencies and propagation constants and with equal amplitudes. The electric fields of both plane waves are oriented in the y direction, which means that they have the same "polarization". The time and space variation of the two electric fields are given by: u: (2.1) = cos(@t - kz) uz(2.1) = cos ([o + Ao]t - [k + Ak]2) We will show that the sum of these two waves produce an intensity envelope in time and space that modulates the carrier frequency of m + Ao/2. The velocity of the envelope is called the group velocity. Assume Ao is << than o. For this problem, consider the sum of two slightly different waves with similar electric fields: v.(z) = sin(ot - kz) vz(z1) = sin([o + Ao]t - [k + Ak]z) Derive analytically an expression for a) the group and b) the phase velocity for the sum of the two fields. HINT: sinfa) +…

Explain the answer to the second and last equation in detail

Use the commutator results, [â, ô] = iħ, [x²,p] = 2iħâ, and [ÂÂ, Ĉ] = Â[Â, Ĉ] + [‚ Ĉ]B, to find the commutators given below. (a) [x, p²] (b) [x³, p] (c) [x², p²]

An electric field in a vacuum is determined to be polarized in the +y-direction. It has a peak of 250 μV/m at z = 0, and the next peak is at z = 5 cm. Determine phase velocity, propagation constant, wavelength, period, frequency, and radial frequency for this wave and write a complete function for e(z,t). Complete Solution please

It has been proposed that spacecraft could be propelled by emitting intense beams of EM radiation from their tails rather than high-velocity chemical exhaust. This could reduce the fuel load required, which is very large for chemical rockets. Suppose a spacecraft (M = 250,000 kg) is floating in empty space at rest relative to a nearby space station. It then emits a very powerful 10.0 MW beam of laser light for one full day from its tail. Calculate the velocity acquired by the vehicle relative to the space station due to this emission of EM radiation

= The electric fields of two harmonic waves of angular frequency w, and we are given by E₁ E₁,0cos (k₁y + w₁t)i and E2 = E2,0cos (k₁y-w₂t)î. Find (a) the instantaneous Poynting vector for the resultant wave motion and (b) the time-average Poynting vector. If the direction of propagation of the second wave is reversed, find (c) the instantaneous Poynting vector for the resultant wave motion and (d) the time -average Poynting vector.

Your video analysis of the motion of a marble gives it position in frame 23 as (x23, 923) = (0.134 m, 0.120 m) and its position in frame 24 as (x24, Y24) = (0.122 m, 0.112 m). You esti- mate that you can measure the x and y positions with uncertainty +0.003 m. The frame rate of the video is 30 frames/s, which means the time interval between frames is At = 0.033 333 s. The uncertainty of the frame rate of a video camera is VERY small. For the sake of this prob- lem, use (At) = 1 x 10-6 s. The mass of the marble is (2.031 +0.001) x 10-2 kg. Calculate the following quantities: 1. The value of the velocity component U = Xi+1Xi At and its uncertainty, Yi+1 Yi = and its uncertainty, At 2. The value of the velocity component vy 3. The value of the velocity magnitude v = 4. The value of the momentum component p 5. The value of the momentum component Py = muy and its uncertainty, and 6. The value of the kinetic energy K = =mv² = m(v²+v²) and its uncertainty. v2+2 and its uncertainty, = mv, and…