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Step-by-Step Solution
Step 1: Identify the Given Data
The widths of the two slits are in the ratio 4:1. Let the intensity from the slit of width 4 units be $I_1$ and that from the slit of width 1 unit be $I_2$. Typically, intensity is proportional to the slit width, so we can write:
$I_1 = 4 I_0 \text{ and } I_2 = I_0.$
Step 2: Express the Resultant Intensity for Maxima
When the two waves interfere constructively, the resultant intensity $I_{\max}$ is given by:
$$
I_{\max} = \bigl(\sqrt{I_1} + \sqrt{I_2}\bigr)^2.
$$
Substituting $I_1 = 4\,I_0$ and $I_2 = I_0$, we get:
$$
I_{\max}
= \Bigl(\sqrt{4\,I_0} + \sqrt{I_0}\Bigr)^2
= \Bigl(2\sqrt{I_0} + \sqrt{I_0}\Bigr)^2
= (3\sqrt{I_0})^2
= 9\,I_0.
$$
Step 3: Express the Resultant Intensity for Minima
When the two waves interfere destructively, the resultant intensity $I_{\min}$ is given by:
$$
I_{\min}
= \bigl(\sqrt{I_1} - \sqrt{I_2}\bigr)^2.
$$
Substituting $I_1 = 4\,I_0$ and $I_2 = I_0$, we get:
$$
I_{\min}
= \Bigl(\sqrt{4\,I_0} - \sqrt{I_0}\Bigr)^2
= \Bigl(2\sqrt{I_0} - \sqrt{I_0}\Bigr)^2
= (\sqrt{I_0})^2
= I_0.
$$
Step 4: Calculate the Ratio of $I_{\max}$ to $I_{\min}$
Finally, the ratio of the maximum intensity to the minimum intensity is:
$$
\frac{I_{\max}}{I_{\min}} = \frac{9\,I_0}{I_0} = 9:1.
$$
