1. Relate slit width to amplitude

   Given: the field strength (amplitude) varies according to slit width.

   If one slit has width $d$ and the other has width $xd$, then their amplitudes are proportional to $a_1 \propto d, \qquad a_2 \propto xd$ So we can write $a_1 = a, \qquad a_2 = xa$

2. Use expressions for maximum and minimum intensity

   In YDSE, if amplitudes are $a_1$ and $a_2$, then $I_{\max} = (a_1+a_2)^2$ $I_{\min} = (a_1-a_2)^2$

   Therefore, $\frac{I_{\max}}{I_{\min}} = \frac{(a_1+a_2)^2}{(a_1-a_2)^2}$

   Substituting $a_1=a$ and $a_2=xa$, $\frac{I_{\max}}{I_{\min}} = \frac{(a+xa)^2}{(a-xa)^2} = \frac{(1+x)^2}{(1-x)^2}$

3. Use the given ratio

   Given $\frac{I_{\max}}{I_{\min}} = \frac{9}{4}$

   So, $\frac{(1+x)^2}{(1-x)^2} = \frac{9}{4}$

   Taking square root, $\frac{1+x}{|1-x|} = \frac{3}{2}$

   Since slit widths are positive and usually $x>1$ for the wider slit, use $|1-x|=x-1$: $\frac{1+x}{x-1} = \frac{3}{2}$

4. Solve for $x$

   $2(1+x)=3(x-1)$ $2+2x=3x-3$ $x=5$

5. Check

   If $x=5$, $\frac{I_{\max}}{I_{\min}}=\frac{(1+5)^2}{(5-1)^2} = \frac{36}{16}=\frac{9}{4}$ which matches.

6. Option check

   • A: $2$ gives $\frac{9}{1} \ne \frac{9}{4}$
   • B: $3$ gives $\frac{16}{4}=4 \ne \frac{9}{4}$
   • C: $4$ gives $\frac{25}{9} \ne \frac{9}{4}$
   • D: $5$ gives $\frac{36}{16}=\frac{9}{4}$ ✅

Therefore, the correct answer is D: $5$.