Answer to Question #276402 in Physics for Rakibul

Question

. (a) Define the interference and diffraction of light.

(b) Write down the difference between the interference and diffraction of light?

(c) Derive an equation for interference of light for bright and dark fringes.

Answer 1

(a) When two (or more) coherent light waves are superimposed, there is a redistribution of the intensity of the waves, resulting in maxima in some places, and the minimums of intensity in others.

This phenomenon is called light interference.

Diffraction is an optical phenomenon associated with a change in the direction of propagation of light waves (compared with the direction provided by the laws of geometric optics) and the spatial redistribution of their intensity under the influence obstacles and inhomogeneities of the environment in their path.

(b) Diffraction of light occurs due to the superposition of secondary waves that generates from various parts of a wavefront. Interference is the result of the superposition of light waves from two coherent sources; intensity of all the bright fringes is not the same in the case of diffraction however it is same in case of interference etc.

(c) 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.