In an interference experiment, the ratio of the intensities of the bright and dark fringes is 16:1. The ratio of the amplitudes due to the two slits is

Options

A. A) `3 : 1`
B. B) `4 : 1`
C. C) `5 : 1`
D. D) `5 : 3`

Answer

To solve the problem, we need to find the ratio of the amplitudes due to the two slits based on the given ratio of the intensities of the bright and dark fringes.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- The ratio of the intensities of the bright fringe (I_max) to the dark fringe (I_min) is given as 16:1.
- This can be expressed mathematically as:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{16}{1} \]

2. **Relating Intensity to Amplitude**:
- The intensity (I) of a wave is proportional to the square of its amplitude (A). This can be expressed as:
\[ I \propto A^2 \]
- Therefore, we can write:
\[ I_{\text{max}} \propto A_{\text{max}}^2 \quad \text{and} \quad I_{\text{min}} \propto A_{\text{min}}^2 \]

3. **Setting Up the Ratio**:
- From the relationship between intensity and amplitude, we can express the ratio of the intensities in terms of the amplitudes:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{A_{\text{max}}^2}{A_{\text{min}}^2} \]
- Substituting the known ratio of intensities:
\[ \frac{16}{1} = \frac{A_{\text{max}}^2}{A_{\text{min}}^2} \]

4. **Taking Square Roots**:
- To find the ratio of the amplitudes, we take the square root of both sides:
\[ \frac{A_{\text{max}}}{A_{\text{min}}} = \sqrt{\frac{16}{1}} = \sqrt{16} = 4 \]

5. **Final Ratio of Amplitudes**:
- Thus, the ratio of the amplitudes due to the two slits is:
\[ \frac{A_{\text{max}}}{A_{\text{min}}} = 4:1 \]

### Conclusion:
The ratio of the amplitudes due to the two slits is \(4:1\). Therefore, the correct option is **(b) 4:1**.