1. For Fraunhofer diffraction at a single slit of width $a$, the intensity distribution is $I(\theta)=I(0)\left(\frac{\sin \beta}{\beta}\right)^2,$ where $\beta=\frac{\pi a\sin\theta}{\lambda}.$

2. At the principal maximum, $\theta=0$, so the intensity is maximum. The amplitude at the center is proportional to the slit width $a$ because all parts of the slit contribute in phase.

3. Hence, $\text{Amplitude at central maximum} \propto a$ and therefore $I_\text{principal} \propto a^2.$

4. If the slit width is doubled: $a \to 2a$ then the amplitude becomes twice, so the intensity becomes $(2)^2=4$ times the original intensity.

Thus, $I' = 4I_0.$

5. Therefore the correct option is:

A: $4I_0$

6. Comparison with stored answer:

• Stored correct answer: D: $I_0$
• Derived answer: A: $4I_0$

The stored answer appears incorrect because the central maximum intensity in single-slit diffraction varies as the square of slit width.