In the fraunhaufer differaction from a single slit illuminated by polychromatic light, the first minimum with wavelength `lamda_(1)` is found to be coincident with the third minimum at `lamda_(2)`. Then find the value of `(lamda_(1))/(lamda_(2))`

Explanation

To solve the problem, we need to analyze the conditions for the minima in a single-slit diffraction pattern. The first minimum for a wavelength \( \lambda_1 \) coincides with the third minimum for a wavelength \( \lambda_2 \).

### Step-by-step Solution:

1. **Understanding the Position of Minima**:
The position of the minima in a single-slit diffraction pattern is given by the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where \( y_n \) is the position of the \( n^{th} \) minimum, \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slit to the screen, and \( d \) is the width of the slit.

2. **Setting Up the Equations**:
- For the first minimum with wavelength \( \lambda_1 \):
\[ y_1 = \frac{1 \cdot \lambda_1 D}{d} \]
- For the third minimum with wavelength \( \lambda_2 \):
\[ y_3 = \frac{3 \cdot \lambda_2 D}{d} \]

3. **Equating the Positions**:
Since the first minimum of \( \lambda_1 \) coincides with the third minimum of \( \lambda_2 \), we can set the two equations equal to each other:
\[ \frac{\lambda_1 D}{d} = \frac{3 \lambda_2 D}{d} \]

4. **Cancelling Common Terms**:
We can cancel \( D \) and \( d \) from both sides (assuming \( D \) and \( d \) are not zero):
\[ \lambda_1 = 3 \lambda_2 \]

5. **Finding the Ratio**:
To find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = 3 \]

### Final Answer:
\[ \frac{\lambda_1}{\lambda_2} = 3 \]