In Young's double-slit experiment, the ratio of intensities of a bright band and a dark band is `16:1`. The ratio of amplitudes of interfering waves will be

Options

A. `16:1`
B. `4:1`
C. `3:1`
D. `5:3`

Answer

To solve the problem, we need to find the ratio of the amplitudes of the interfering waves in Young's double-slit experiment given that the ratio of intensities of a bright band to a dark band is 16:1.

### Step-by-step Solution:

1. **Understanding the Relationship Between Intensity and Amplitude**:
In Young's double-slit experiment, the intensity (I) of the light is related to the amplitude (A) of the waves by the formula:
\[ I \propto A^2 \]
This means that the intensity of the bright band (I_max) is proportional to the square of the sum of the amplitudes of the two waves, and the intensity of the dark band (I_min) is proportional to the square of the difference of the amplitudes.

2. **Setting Up the Ratios**:
Let the amplitudes of the two waves be \( A_1 \) and \( A_2 \). The intensity of the bright band can be expressed as:
\[ I_{max} = (A_1 + A_2)^2 \]
The intensity of the dark band can be expressed as:
\[ I_{min} = (A_1 - A_2)^2 \]
Given that the ratio of the intensities is \( \frac{I_{max}}{I_{min}} = \frac{16}{1} \), we can write:
\[ \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = 16 \]

3. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ \frac{A_1 + A_2}{A_1 - A_2} = 4 \]

4. **Cross Multiplying**:
Cross multiplying gives:
\[ A_1 + A_2 = 4(A_1 - A_2) \]

5. **Expanding the Equation**:
Expanding the right side:
\[ A_1 + A_2 = 4A_1 - 4A_2 \]

6. **Rearranging the Terms**:
Rearranging the equation to isolate terms involving \( A_1 \) and \( A_2 \):
\[ A_1 + A_2 + 4A_2 = 4A_1 \]
\[ A_1 - 4A_1 + 5A_2 = 0 \]
\[ -3A_1 + 5A_2 = 0 \]

7. **Solving for the Ratio**:
From the equation \( 3A_1 = 5A_2 \), we can express the ratio of the amplitudes:
\[ \frac{A_1}{A_2} = \frac{5}{3} \]

8. **Final Answer**:
Thus, the ratio of the amplitudes \( A_1 : A_2 \) is \( 5 : 3 \).