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Step-by-Step Detailed Solution
Step 1: Understand the Given Information
We are told that in a Young's double slit experiment, one slit is three times wider than the other. Let the widths of the two slits be:
Width of first slit = b_1 = 3b_2
Width of second slit = b_2
Since the amplitude of light from each slit is proportional to its width, we have:
Amplitude from first slit, A_1 \propto b_1 = 3b_2
Amplitude from second slit, A_2 \propto b_2
Step 2: Relate Amplitudes to Intensities
Intensity of light is proportional to the square of its amplitude. Therefore,
I_1 \propto A_1^2 and I_2 \propto A_2^2.
Since A_1 = 3A_2 (due to widths being in the ratio 3:1), we have:
I_1 = (3A_2)^2 = 9A_2^2.
If I_2 = A_2^2, then
I_1 = 9 I_2.
Step 3: Obtain the Min/Max Intensity Formula
In an interference pattern, the maximum and minimum intensities occur due to constructive and destructive interference, respectively. The intensities are given by:
I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2, \\
I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2.
Hence, the ratio
\frac{I_{\min}}{I_{\max}} = \left(\frac{\sqrt{I_1} - \sqrt{I_2}}{\sqrt{I_1} + \sqrt{I_2}}\right)^2.
Step 4: Substitute the Given Ratio I_1 = 9I_2
Let I_2 = I_2 and I_1 = 9 I_2. Then,
\sqrt{I_1} = \sqrt{9I_2} = 3\sqrt{I_2}, \quad \sqrt{I_2} = \sqrt{I_2}.
Therefore,
\frac{I_{\min}}{I_{\max}} = \left(\frac{3\sqrt{I_2} - \sqrt{I_2}}{3\sqrt{I_2} + \sqrt{I_2}}\right)^2
= \left(\frac{2}{4}\right)^2
= \left(\frac{1}{2}\right)^2
= \frac{1}{4}.
Step 5: Compare with the Required Form
The question states that the ratio is x : 4, which means:
\frac{I_{\min}}{I_{\max}} = \frac{x}{4}.
Comparing this with the value we obtained ( \frac{1}{4} ), we see that
\frac{x}{4} = \frac{1}{4} \implies x = 1.
Final Answer:
The value of x is 1.
