(In a double slit expriment, the two cohunt beams
have slightly different intensities I and I + lI (lI << I). Show that the resultant intersity at the maxima is nearly
4I while that at minima is (dI)^2/4I
-
UP 0 DOWN 0 0 1
1 Answers
1In the superposition of two waves let y1=a sin\frac{2\pi }{\lambda } (vt) & y2=a sin\frac{2\pi }{\lambda } (vt+x) where is the path difference.
If y is the net displacement of the particle, according to principle of superposition,
y=y1+y2
y=a sin\frac{2\pi }{\lambda } (vt)+a sin\frac{2\pi }{\lambda } (vt+x)
then using sinA + sinB=2sin \left((A+B)/2 \right)cos\left((A-B)/2 \right) we get
y= 2a cos\frac{\pi x}{\lambda } sin\frac{2\pi }{\lambda }\left(vt+\frac{x}{2} \right)
y=Rsin\frac{2\pi }{\lambda }\left(vt+\frac{x}{2} \right)
This also represents S.H.M. having amplitude,
R=2a cos \frac{\pi x}{\lambda }
Since intensity, I directly proportional to R2
I = K 4a2cos2\frac{\pi x}{\lambda }
Imax = K.4a2 X 1=4Ka2
Intensity i due to a single source,i = Ka2
So, Imax = 4i
similarly I min =0