In the Young's double slit experiment the central maxima is observed to be ${I_0}$. If one of the slits is covered, then intensity at the central maxima will become-
(A) $\dfrac{{{I_0}}}{2}$
(B) $\dfrac{{{I_0}}}{{\sqrt 2 }}$
(C) $\dfrac{{{I_0}}}{4}$
(D) ${I_0}$

Hint:In this question, the concept of the double slit experiment will be used; it defines the particle nature as well as the wave nature of light. We know that the maximum intensity point is known as central maxima in a double slit experiment. Since the one slit is covered then the intensity at both the slits will be the same which means intensity will be equal at both slits.

Complete step by step answer:

As given in the question that In the Young's double slit experiment the central maxima is observed to be ${I_0}$ and one of the slits is covered. We need to calculate the central maxima intensity for that case.
As we know that the diffraction pattern and interference pattern are obtained by double slit experiment. Diffraction patterns can be obtained from a single slit as well. In the given question one slit is covered so we can write,
${I_1} = {I_2}$
Here, ${I_1}$ and ${I_2}$ are intensities at both the slits respectively.
Let us assume that the common intensity to be $I$.
As we know that central maxima can be calculated as,
${I_o} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} ......\left( 1 \right)$
Now, we substitute $I$ for ${I_1}$ and $I$ for ${I_2}$ in the above equation.
$ \Rightarrow {I_o} = I + I + 2\sqrt {I \cdot I} $
Now, we solve above equation as
$ \Rightarrow {I_o} = I + I + 2\sqrt {{I^2}} $
After simplification of the above equation we get,
$ \Rightarrow {I_o} = I + I + 2I$
Now, we add the terms to obtain,
\[ \Rightarrow I = \dfrac{{{I_o}}}{4}\]

Therefore, the correct option is (C).

Note:As we know that in the double slit experiment the central maxima is obtained at maximum intensity point. The value of intensity can never be zero and remember that one slit is covered so the value of intensity for the covered slit will be zero. If the value of intensity for the covered slit is considered other than zero, then the solution will be error.