In a Youngis double slits experment, the radis of amplitude of light coming from stits is `2:1`.The ratio of the maximum to minimum intensity in the interference pattern is

Options

A. `9:1`
B. `9:4`
C. `2:1`
D. `25:9`

Answer

To solve the problem regarding the Young's double slit experiment with the given ratio of amplitudes, we will follow these steps:

### Step 1: Understand the given data
We are given the ratio of amplitudes of light coming from the two slits. Let:
- Amplitude from slit 1, \( A_1 = 2a \)
- Amplitude from slit 2, \( A_2 = a \)

### Step 2: Write the formulas for maximum and minimum intensity
In an interference pattern:
- The maximum intensity \( I_{\text{max}} \) is given by:
\[ I_{\text{max}} = (A_1 + A_2)^2 \]
- The minimum intensity \( I_{\text{min}} \) is given by:
\[ I_{\text{min}} = (A_1 - A_2)^2 \]

### Step 3: Substitute the values of amplitudes
Substituting the values of \( A_1 \) and \( A_2 \):
- For maximum intensity:
\[ I_{\text{max}} = (2a + a)^2 = (3a)^2 = 9a^2 \]
- For minimum intensity:
\[ I_{\text{min}} = (2a - a)^2 = (a)^2 = a^2 \]

### Step 4: Calculate the ratio of maximum to minimum intensity
Now, we find the ratio \( \frac{I_{\text{max}}}{I_{\text{min}}} \):
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9a^2}{a^2} = 9 \]

### Conclusion
Thus, the ratio of maximum to minimum intensity in the interference pattern is:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = 9:1 \]

### Final Answer
The ratio of maximum to minimum intensity is \( 9:1 \).

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