Describe the condition for constructive and destructive interfernce.

Explanation

Conditions for constructive and destructive interference : Let the two slits `S_(1)` and `S_(2)` are illuminated by the monochromatic light from a single light source `lambda` = Wavelength of length emitted by source S.
Let a, b be the amplitude of two light waves emitted from `S_(1)` and `S_(2)` respectively.
`psi`= Phase difference by which the wave emitted from `S_(1)` leads the wave emitted from `S_(2)`
Two waves emitted by `S_(1)` and `S_(2)` are represented by:
`y=y_(1) + y_(2)`
`y = a sin omegat + b sin (omega t + phi)`
`y = sin omegat (a + b cos phi) + b cos omega t sin phi`....(i)
Let `a + b cos phi = R cos theta`......(ii)
`b sin phi = R sin theta`.......(iii)

where R= Amplitude of resultant wave and `theta`= Phase difference between resultant wave and wave emitted from source `S_(1)` From eqn. (i) we get
`y =R sin omega t cos theta + R cos omega t sin theta`
`y= R sin (omega t + theta)`
Resultant wave is a simple harmonic wave having amplitude R.
Thus, resultant wave is a simple harmonic wave having amplitude R and phase difference `theta` with respect to the wave emitted by `S_(1)`.
Step 2. Amplitude of Resultant Wave Squaring and adding equation (i) and (ii) we get,
`(a + b cos phi)^(2) + b^(2) sin^(2)phi = R^(2) cos^(2)theta + R^(2) sin^(2) theta`
`R^(2) = a^(2) + b^(2) + 2ab cos phi`.....(iv)
`R = sqrt(a^(2) + b^(2) + 2ab cos hi)`
Step 3. Now, Resultant Intensity `(I_(R))` at `P prop ("Amplitude")^(2)` of the resultant wave.
`I_(R) = KR^(2) =K(a^(2) +b^(2)) + 2ab cos phi`.....(v)
Dividing (iii), (iv) , we get
`tan theta = (b sin phi)/(a + b cos phi)` or `theta = tan^(-1) (b sin phi)/(a + b cos phi)`
Dividing (iii) by (iv), we get
`tan theta = (b sin phi)/(a + b cos phi)` or `theta = tan^(-1) (b sin phi)/( a + b cos phi)`
Again intensities of individual waves can be given as
`rArr I_(a) = Ka^(2)` and `I_(b) = Kb^(2)`...(vi)
`I_(a) prop a^(2)` and `I_(b) prop b^(2)`......(vii)
Using equation {vii) in relation (v), we get
`I_(R) = (I_(a) + I_(b) +2 sqrt(I_(a) I_(b) cos phi)`
Clearly, the resultant intensity due to interference is not simple sum of intensities due to individual waves i.e. `I_(a) + I_(b)` but has another additional factor `(2 sqrt(I_(a)I_(b) cos phi)` {also which is known as interference factor).
Conditions for Constructive and Destructive Interference Constructive Interference
From eqn. (v),
`phi = 0, 2pi, 4pi`,.......
Intensity will be maximum at those points on the screen where path difference between the two interfering waves is an integral multiple of wave length. These points are called, points of constructive interference or interference maxima.
Intensity at the position of constructive interference will be maximum and the relation for the. same is given by:
`= sqrt(I_(a)) + sqrt(I_(b))^(2)`
Destructive Interference
For destructive. interference,, intensity (`I_(R)`) is minimum.
From eqn. (v)
`I_(R) = I_("min")` If `cos phi `= Minimum = - 1
or `phi = pi, 3pi, 5pi`,.....
In general, `phi = (2n+1) phi, n =0,1,2,`.......
Thus, destructive interference takes place if the phase difference between the two interfering waves is an odd integral multiple of `pi`.
or path difference `Deltax = (2n+1) lambda/2, n =0,1,2,3`,...
So for destructive interference, path difference between the two interfering waves is an odd integral multiple, of `lambda`.