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**Hint:**We must know the formula and the various conditions of Fresnel diffraction for solving this question. Therefore, upon knowing the conditions and formula for Fresnel diffraction, the solution will just be a formula based solution.

**Complete step by step solution:**

In Fresnel diffraction, the distance between the source (slit) and the screen is assumed to be very much larger than the distance of the image from the central bright fringe/spot. As a result, the viewing angle is assumed to be very small. The second assumption made in Fresnel diffraction is that the size of the slit is assumed to be comparable to the distance of the image from the central bright fringe/spot. We can write down these conditions numerically:

$d\gg {{x}_{i}}$,

where $d$ is the distance between the source and the screen and ${{x}_{i}}$ is the distance if the image formed from the central bright fringe/spot.

$\Rightarrow \theta $ is very small,

where, $\theta $ is the viewing angle.

$a$ is comparable to ${{x}_{i}}$,

where, a is the width of the slit.

The formula for Fresnel diffraction is:

$A=n\pi \lambda d$, where $n$ is the number of blocked half-period zones(HPZ).

Therefore, $\Rightarrow nd=\dfrac{A}{\pi \lambda }$

We know that $A$, $\pi $ and $\lambda $ are constants.

Therefore, $nd=constant$ .

Hence, $n\propto \dfrac{1}{d}$.

If $d$ is decreased, the number of blocked HPZ will be increased as they are inversely proportional to each other.

Since, the number of blocked HPZ is also inversely proportional to the intensity, i.e.:

$I\propto \dfrac{1}{n}$

Therefore, $I$ decreases as $n$ increases.

Hence, the intensity of the central bright fringe/spot will decrease upon the decrease in the distance between the disc and the screen.

**Therefore, option (B) is correct.**

**Note:**There are two main types of diffraction for fringes: one is Fresnel diffraction (as discussed above), and the other is Fraunhofer diffraction. There is not much difference between the two, but we must not confuse one with the other.

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