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Step-by-Step Solution
Step 1: Write the expressions for maximum and minimum intensities
The maximum intensity, denoted by $I_{\max}$, occurs when the crests (or troughs) of the two waves coincide constructively. It is given by:
$$
I_{\max} = \bigl(\sqrt{I_1} + \sqrt{I_2}\bigr)^2
$$
The minimum intensity, denoted by $I_{\min}$, occurs when the crest of one wave coincides with the trough of the other, producing destructive interference. It is given by:
$$
I_{\min} = \bigl(\sqrt{I_1} - \sqrt{I_2}\bigr)^2
$$
Step 2: Use the given ratio of intensities
We are given that the ratio of the two source intensities is
$$
\frac{I_1}{I_2} = n \quad\Longrightarrow\quad I_1 = n\, I_2.
$$
Step 3: Express the ratio $I_{\max}/I_{\min}$ using $n$
Divide $I_{\max}$ by $I_{\min}$:
$$
\frac{I_{\max}}{I_{\min}}
= \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}
= \biggl(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\biggr)^2.
$$
Substitute $I_1 = n\, I_2$:
$$
\frac{I_{\max}}{I_{\min}}
= \biggl(\frac{\sqrt{n\,I_2} + \sqrt{I_2}}{\sqrt{n\,I_2} - \sqrt{I_2}}\biggr)^2
= \biggl(\frac{\sqrt{n} + 1}{\sqrt{n} - 1}\biggr)^2.
$$
Step 4: Express the desired ratio $ \displaystyle \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} $
We employ the identity
$$
\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}
= \frac{\frac{I_{\max}}{I_{\min}} - 1}{\frac{I_{\max}}{I_{\min}} + 1}.
$$
So let
$$
R = \frac{I_{\max}}{I_{\min}}
= \biggl(\frac{\sqrt{n} + 1}{\sqrt{n} - 1}\biggr)^2.
$$
Then
$$
\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}
= \frac{R - 1}{R + 1}.
$$
Step 5: Simplify the expression
First, notice that
$$
\biggl(\frac{\sqrt{n} + 1}{\sqrt{n} - 1}\biggr)^2
= \frac{(\sqrt{n} + 1)^2}{(\sqrt{n} - 1)^2}.
$$
Now compute:
$$
(\sqrt{n} + 1)^2 = n + 2\sqrt{n} + 1,
\quad (\sqrt{n} - 1)^2 = n - 2\sqrt{n} + 1.
$$
Hence,
$$
R - 1
= \frac{n + 2\sqrt{n} + 1}{n - 2\sqrt{n} + 1} - 1
= \frac{n + 2\sqrt{n} + 1 - (n - 2\sqrt{n} + 1)}{n - 2\sqrt{n} + 1}
= \frac{4\sqrt{n}}{n - 2\sqrt{n} + 1}.
$$
Similarly,
$$
R + 1
= \frac{n + 2\sqrt{n} + 1}{n - 2\sqrt{n} + 1} + 1
= \frac{n + 2\sqrt{n} + 1 + (n - 2\sqrt{n} + 1)}{n - 2\sqrt{n} + 1}
= \frac{2n + 2}{n - 2\sqrt{n} + 1}.
$$
Thus,
$$
\frac{R - 1}{R + 1}
= \frac{4\sqrt{n}}{n - 2\sqrt{n} + 1} \,\bigg/\, \frac{2n + 2}{n - 2\sqrt{n} + 1}
= \frac{4\sqrt{n}}{2(n + 1)}
= \frac{2\sqrt{n}}{n + 1}.
$$
Therefore, the ratio is
$$
\frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{2 \sqrt{n}}{n + 1}.
$$
