Question

An interference with two coherent sources the fringe width varies
(A). Directly as wavelength.
(B). Inversely as wavelength.
(C). Directly as separation between slits.
(D). Inversely as the distance between slit and screen.

Answer 1

Hint: In physics interference is a very important phenomenon. Interference involves the superimposition of two waves which forms a resultant with amplitude more, less or the same as the two waves involved. There are also two specific conditions that are important for us – constructive and destructive interference. For constructive and destructive interference it is very important that the two waves either come from the same source or have the same or nearly the same frequency. Interference occurs in all types of waves including but not limited to acoustic, surface water waves, gravity waves, matter water and light waves.

Complete step-by-step answer:
In interference the amplitude of the resultant wave, is a vector sum of the amplitude of the two waves. If a crest of one wave meets a crest of another wave, then the amplitude is the sum of amplitude of the two waves. If a crest of one wave meets the trough of another waves then the amplitude of the resultant wave is the difference in the amplitude of the two waves.
We know that the fringe width is given by
$X = \dfrac{{\lambda D}}{d}$
$X = $Fringe width
$D = $Distance between the source and the screen
$d = $Distance between the two sources
$\lambda = $Wavelength of the light used
Here, we can clearly see that
$X \propto \lambda $
Hence, option A is the correct choice.

Note: As discussed earlier to produce constructive and destructive interference, we need two waves from the same source or with constant phase difference. In labs, white light is not usually used to demonstrate interference, normally we use sodium lamps as a single light source for the two waves.