In Young’s interference experiment, the central bright fringe can be identified due to the fact that it
A. has greater intensity than other fringes which are bright.
B. is wider than the other bright fringes.
C. is narrower than the other bright fringes.
D. can be obtained by using white light instead of monochromatic light.

Hint: In this question, we need to comment on the fact on which the central bright fringe in the Young’s interference experiment could be identified. For the central bright fringe put \[n = 0\] in the fringe position formula and relate the things with the given options.

Complete step by step answer:
The position of the fringe is given by $y = \dfrac{{n\lambda D}}{d}$
And the fringe width is given by the formula $\beta = \dfrac{{\lambda D}}{d}$
Now for the central bright fringe we put \[n = 0\]
Thus the formula for the position becomes $y = 0$
The central bright fringe doesn’t depend on the wavelength of light. However for $n \ne 0$ the position of the other bright fringes depends on the wavelength of light used. Therefore for all light of different wavelengths the central bright fringe is the same but the position of the other fringes differs depending upon the wavelength of light.
Thus in Young’s interference experiment, the central bright fringe can be identified due to the fact that it can be obtained by using white light instead of monochromatic light.

So, the correct answer is “Option D”.

Note:
The Young’s interference experiment is a double slit experiment. The results will be different for a single slit experiment. Also, students need to remember that monochromatic light means light of only a single wavelength while white light consists of all seven colours and thus different wavelengths.