Question

P4)
A material having an index of refraction of 1.40 is used as an antireflective coating on a piece of glass (n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 590 nm light?
nm

P6)

One radio transmitter A operating at 56.5 MHz is 10.6 m from another similar transmitter B that is 180° out of phase with A. How far must an observer move from A toward B along the line connecting A and B to reach the nearest point where the two beams are in phase?

Answer 1

P4)     t = λ / 4n
= 590 / 4*1.4
= 105.357 nm

P5) λ = c/f = 3x10⁸ / 56.5 x10⁶ = 5.3097345m

AB is 10λ/5.3097345 so a standing wave, length 1.8833333λ, is set-up between A and B,

Consider what happens at each end, The waves will meet out of phase, so there will be a node at each end. Node to antinode distance = λ/4. The pattern of nodes (A) and antinodes (N) will be
N-A-N-A-N-A-N-A-N

Antinodes are where the 2 beams are in phase.

So from end A we must move a distance λ/4 to get to the nearest point where the two beams are in phase (the first antinode). So the answer is 5.3097345/4 = 1.32743363m