For minima to take place between two mono-chromatic light waves of wavelength $\lambda $, the path difference should be
A) $n\lambda $
B) $(2n - 1)\dfrac{\lambda }{4}$
C) $(2n - 1)\dfrac{\lambda }{2}$
D) $(2n - 1)\lambda $

Hint:A light that has only one wavelength is known as monochromatic light. The electromagnetic radiation of light is not purely monochromatic. It has a wavelength of very short range. The wavelength of these radiations range from picometres to kilometres.

Step-By-Step Explanation:
Step I:
When two waves superimpose with each other then the resultant wave formed either has greater or lower amplitude. This phenomenon is known as interference of light. When two waves have the same frequencies and travel in a medium the resultant wave at different points is negligible and it is called destructive interference.
Step II:
In destructive interference, two waves cancel out each other and there is no wave left. The path difference is defined as the difference in the path travelled by two waves. The path difference between two destructive waves is an odd multiple of $\dfrac{\lambda }{2}$. Hence, if two monochromatic light waves have wavelength $\lambda $ then the formula used for the path difference between two destructive waves can be given by,
$\Delta = (2n - 1)\dfrac{\lambda }{2}$

Step III:
Therefore, the path difference between two monochromatic waves of light should be \[(2n - 1)\dfrac{\lambda }{2}\]

$ \Rightarrow $ Option C is the correct answer.

Note:It is important to note that in case of destructive interference, the waves have amplitude in opposite directions. In this phenomenon, the crest of one wave coincides with the trough of another wave. This reduces the intensity of the resultant wave. At minima, the pattern of the fringes obtained will be dark. The path difference is responsible for the constructive or destructive interference of waves.