Question

Derive an equation for interference of light for bright and dark fringes

Answer 1

When superimposing two coherent waves

$E_1=E_{01}\cos (\omega t-\frac{2\pi}{\lambda }r_1+\alpha_1)$

$E_2=E_{02}\cos (\omega t-\frac{2\pi}{\lambda }r_2+\alpha_2)$

the intensity of the resulting wave is equal to $I$ ~ $E^2$ . So, we get

...

$I=I_1+I_2+2\sqrt{I_1I_2}\cos\Delta\phi$ .

The equation shows that at points in space where $\cos\Delta\phi>0,\ I>I_1+I_2$ , ie there is an increase in intensity (maximums). If $\cos\Delta\phi<0,\ I<I_1+I_2$ . In this case there is a decrease in light intensity (minimums).

The wave intensity will be maximum at $\cos\Delta\phi=1$ . Then:

$\Delta \phi=\pm2\pi k,\ \Delta=\pm k\lambda\ (k=0,1,2, ...)$ .

The minimum intensity will correspond to $\cos\Delta\phi=-1$ . In this case:

$\Delta \phi=\pm(2k+1)\pi,\ \Delta=\pm (2k+1)\frac{\lambda}{2}\ (k=0,1,2, ...)$ ,

where $\Delta$ is the optical path length.