A) as it a narrower than other bright fringes
B) by using white light instead of monochromatic light
C) as it has a greater intensity than other bright fringes
D) as it is wider than other bright fringes.
Correct Answer: B
Solution :
Interference fringes are always equally spaced & also have same intensity \[({{I}_{\max }})\] as central part is always bright. If sources are given white light which consists of a no. of wavelength (colour). Then light of each wavelength gives its own set of fringes & hence fringe width will be different for different\[\lambda .\] Only the central (zero order) fringes of all wavelength lie in same positions for each colour. The fringes of different colours are therefore intermixed. Hence only a few coloured fringes are obtained when path difference is large. There is a very much intermixing of colours at any point so that the resultant colour is white & uniform illumination is obtained. Fringe pattern will disappear. Hence either the light should be monochromatic or the path difference between the interfering waves must be small. Hence in our options for Ans. none is correct, is in suffficient to be correct.
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