In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be:

Options

A. `(sqrt3 + 1)^(4) : 16`
B. `25 :9`
C. `4 : 1`
D. `9 : 1`

Explanation

To solve the problem, we need to find the ratio of the intensity of maxima to minima in a Young's double slit experiment where the ratio of the slit widths is given as 4:1.

### Step-by-Step Solution:

1. **Identify the slit widths**: Let the widths of the two slits be \( W_1 \) and \( W_2 \). Given the ratio \( W_1 : W_2 = 4 : 1 \), we can express this as:
\[ W_1 = 4k \quad \text{and} \quad W_2 = k \]
for some constant \( k \).

2. **Relate intensity to slit width**: The intensity \( I \) produced by a slit is directly proportional to the square of its width. Therefore, we can express the intensities as:
\[ I_1 \propto W_1^2 \quad \text{and} \quad I_2 \propto W_2^2 \]
This gives us:
\[ I_1 = C(4k)^2 = 16Ck^2 \quad \text{and} \quad I_2 = C(k)^2 = Ck^2 \]
where \( C \) is a constant of proportionality.

3. **Calculate the ratio of intensities**: The ratio of the intensities is:
\[ \frac{I_1}{I_2} = \frac{16Ck^2}{Ck^2} = 16 \]
Thus, we have \( I_1 : I_2 = 16 : 1 \).

4. **Calculate the intensity of maxima**: The intensity of the maxima \( I_{max} \) is given by:
\[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \]
Substituting the values:
\[ I_{max} = (\sqrt{16} + \sqrt{1})^2 = (4 + 1)^2 = 5^2 = 25 \]

5. **Calculate the intensity of minima**: The intensity of the minima \( I_{min} \) is given by:
\[ I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \]
Substituting the values:
\[ I_{min} = (\sqrt{16} - \sqrt{1})^2 = (4 - 1)^2 = 3^2 = 9 \]

6. **Find the ratio of intensities**: Now, we can find the ratio of \( I_{max} \) to \( I_{min} \):
\[ \frac{I_{max}}{I_{min}} = \frac{25}{9} \]

### Final Answer:
The ratio of the intensity of maxima to minima is \( \frac{25}{9} \).