. The resultant minimum intensity is, a a I I min min ∝ ( . ) As a special case, if a = a = a , then equation ( . ).
becomes, a a c a c os os ) a cos ( / ) a = cos( / ) ( . ) a ∝ ∝ cos ( / ) ( . ) a ∝ I cos ( / ) ( . ) max = when, f p p ± , , ..., ( .
) I min = when, f p p p = ± ± ± , , ..., ( . ) We conclude that the phase difference ϕ , between the two waves decides the intensity of light at that point where the two waves meet. EXAMPLE . Two light sources with amplitudes units and units respectively interfere with each other.
Calculate the ratio of maximum and minimum intensities. Solution Amplitudes, a = , a = Resultant amplitude, a a a a cos f The two waves have different amplitudes a and a , same angular frequency ω , and a phase difference of ϕ between them. The resultant displacement will be given by, y = y + y = a sin ωt + a sin ( ωt + ϕ ) ( . ) The simplification of the above equation by using trigonometric identities as done in (XI Physics .
) gives, y = A sin ( ωt + θ ) ( . ) Where, A a a a a cos f ( . ) q tan sin cos a a a ( . ) The resultant amplitude is maximum, a a max = ; when ϕ = , ± π , ± π .
. . , ( . ) The resultant amplitude is minimum, a a min = ; when ϕ = ± π , ± π , ± π ..., ( .
) The intensity of light is proportional to square of amplitude, I ∝ A ( . ) Now, squaring equation ( . ) on both sides, I