Question

(a) Write the conditions under which two light waves originating from two coherent sources can interfere each other (i) constructively, and (ii) destructively, in terms of wavelength. Can these be applied for two lights originating from two sodium lamps ? Give reason.

(b) Monochromatic light of green colour is used in Young's double slit experiment and an interference pattern is observed on a screen. If the green light is replaced by red monochromatic light of the same intensity, how will the fringe width of interference pattern be affected ? Justify your answer.

Hint:

Constructive: path difference n*lambda; destructive: (n + 1/2)*lambda; beta = lambda*D/d.

Answer 1

Step 1: Conditions for interference (a):
Two coherent sources keep a constant phase difference. At a point, the resultant intensity depends on the path difference \(\Delta x\) between the two waves.
(i) Constructive interference (bright): the waves arrive in phase, so \(\Delta x = n\lambda\), where \(n = 0, 1, 2, \ldots\) (a whole number of wavelengths).
(ii) Destructive interference (dark): the waves arrive in opposite phase, so \(\Delta x = \left(n + \dfrac{1}{2}\right)\lambda\) (an odd number of half-wavelengths).

Step 2: Two sodium lamps (a):
No, these conditions cannot be applied. Two separate sodium lamps are not coherent sources.
The atoms in each lamp emit light independently, and the phase of the emitted light changes randomly about every \(10^{-10}\) s. So the phase difference between the two lamps keeps changing, the bright and dark positions keep shifting, and the eye sees only uniform illumination.

Step 3: Fringe width (b):
In Young's experiment the fringe width is \(\beta = \dfrac{\lambda D}{d}\), where \(D\) is the slit-to-screen distance and \(d\) is the slit separation.
Red light has a longer wavelength than green light (about 650 nm against 530 nm). \(D\) and \(d\) are unchanged, so \(\beta \propto \lambda\).
So the fringe width increases when green light is replaced by red light. The intensity does not appear in the formula, so keeping it the same has no effect on the fringe width.

Final Answer:
(a) Constructive: \(\Delta x = n\lambda\); destructive: \(\Delta x = (n + \tfrac{1}{2})\lambda\). They do not apply to two sodium lamps because the lamps are not coherent. (b) The fringe width increases, as \(\beta \propto \lambda\) and \(\lambda_{red} > \lambda_{green}\).\[ \boxed{\beta = \dfrac{\lambda D}{d} \ \Rightarrow \ \beta_{red} > \beta_{green}} \]