1. Given

   • Widths of the two slits are in the ratio $w_1:w_2=9:16$
   • Intensity is directly proportional to slit width, so $I_1:I_2=9:16$

2. Relation for maximum and minimum intensity in interference If the individual intensities are $I_1$ and $I_2$, then $I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2$ $I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2$

3. Substitute the ratio Let $I_1=9k, \quad I_2=16k$ Then $\sqrt{I_1}=3\sqrt{k}, \quad \sqrt{I_2}=4\sqrt{k}$

4. Compute maximum intensity $I_{\max}=(3\sqrt{k}+4\sqrt{k})^2=(7\sqrt{k})^2=49k$

5. Compute minimum intensity $I_{\min}=(4\sqrt{k}-3\sqrt{k})^2=(\sqrt{k})^2=k$

6. Find the ratio $\frac{I_{\max}}{I_{\min}}=\frac{49k}{k}=49$ Hence, $I_{\max}:I_{\min}=49:1$

7. Option check

   • A: $3:4$ ❌
   • B: $4:3$ ❌
   • C: $7:1$ ❌
   • D: $49:1$ ✅

Therefore, the correct answer is D.