In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Options

A. `1:4`
B. `3:1`
C. `4:1`
D. `2:1`

Answer

To solve the problem, we need to analyze the amplitudes of the light waves coming from the two slits and then calculate the resulting intensities at the points of constructive and destructive interference.

### Step 1: Define the slit widths
Let the width of the first slit (slit 1) be \( D \). According to the problem, the width of the second slit (slit 2) is three times that of the first slit. Therefore, the width of slit 2 is:
\[ d = 3D \]

### Step 2: Determine the amplitudes
The amplitude of the light wave coming from a slit is proportional to its width. Thus, the amplitude \( A_1 \) from slit 1 is:
\[ A_1 \propto D \]
And the amplitude \( A_2 \) from slit 2 is:
\[ A_2 \propto 3D \]

### Step 3: Express the amplitudes in terms of a constant
We can express the amplitudes as:
\[ A_1 = kD \quad \text{and} \quad A_2 = k(3D) = 3kD \]
where \( k \) is a proportionality constant.

### Step 4: Calculate the resultant amplitude
The resultant amplitude \( A \) when two waves interfere is given by the vector sum of the individual amplitudes. The maximum resultant amplitude occurs when the waves are in phase:
\[ A_{\text{max}} = A_1 + A_2 = kD + 3kD = 4kD \]

The minimum resultant amplitude occurs when the waves are out of phase:
\[ A_{\text{min}} = |A_1 - A_2| = |kD - 3kD| = | -2kD | = 2kD \]

### Step 5: Calculate the intensities
The intensity \( I \) of a wave is proportional to the square of its amplitude:
\[ I \propto A^2 \]

Thus, the maximum intensity \( I_{\text{max}} \) is:
\[ I_{\text{max}} \propto (A_{\text{max}})^2 = (4kD)^2 = 16k^2D^2 \]

And the minimum intensity \( I_{\text{min}} \) is:
\[ I_{\text{min}} \propto (A_{\text{min}})^2 = (2kD)^2 = 4k^2D^2 \]

### Step 6: Find the ratio of maximum to minimum intensity
Now, we can find the ratio of maximum intensity to minimum intensity:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{16k^2D^2}{4k^2D^2} = \frac{16}{4} = 4 \]

### Final Answer
The ratio of the maximum to the minimum intensity in the interference pattern is:
\[ \boxed{4} \]