Light waves from two coherent sources superimpose at a point. The waves, at this point, can be expressed as `y_(1) = a sin [10^(15) pi t]` and `y_(2)a sin [10^(15) pi t + phi]`. Find the resultant amplitude if phase difference `phi` is
(a) zero
(b) `pi//3`
(c) `pi`
Also find the frequency (Hz) of resultant wave in each case.

Explanation

Resultant amplitude can be obtained from the relation `A^(2) = A_(1)^(2) + A_(2)^(2) + 2A_(1)A_(2)cos phi`
`[I = I_(1) + I_(2) + 2 sqrt I_(1),I_(2) cos phi]`
`A_(1)` and `A_(2)` are the amplitudes of interfering waves, `phi` is the phase difference at the given point and A is the resultant amplitude. Here, `A_(1) = a` and `A_(2) = 2a`.
(a) `phi = 0, A^(2) = a^(2) + 4a^(2) + 2a(2a) cos 0^(@) = 9a^(2) implies A = 3a`
(b) `phi = phi//3, A^(2) = a^(2) + 4a^(2) + 2a(2a) cos pi//3 = 7a^(2) implies A = sqrt 7a`
(c) `phi = pi, A^(2) = a^(2) + 4a^(2) + 2a(2a) cos pi = a^(2) implies A = a`
Interference results due to superposition of waves of same frequency and frequency of the resultant wave also has the same value. Here, this frequency is
`omega = 10^(15) pi or 2pi f = 10^(15)pi`
`:. f = 5 xx 10^(15) Hz`
.