In young's experiment, the separation between 5th maxima and 3rd minima is how many times as that of fringe width ?

Options

A. 5 times
B. 3 times
C. 2.5 times
D. 2 times

Answer

To solve the problem of finding the separation between the 5th maxima and the 3rd minima in Young's double-slit experiment in relation to the fringe width, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Positions of Maxima and Minima:**
- The position of the nth maxima (bright fringe) is given by the formula:
\[ y_m = \frac{m \lambda D}{d} \]
where \(m\) is the order of the maxima, \(\lambda\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the separation between the slits.

- The position of the nth minima (dark fringe) is given by the formula:
\[ y_n = \frac{(n - 0.5) \lambda D}{d} \]
where \(n\) is the order of the minima.

2. **Calculating the Position of the 5th Maxima:**
- For the 5th maxima (\(m = 5\)):
\[ y_5 = \frac{5 \lambda D}{d} \]

3. **Calculating the Position of the 3rd Minima:**
- For the 3rd minima (\(n = 3\)):
\[ y_3 = \frac{(3 - 0.5) \lambda D}{d} = \frac{2.5 \lambda D}{d} \]

4. **Finding the Separation (p) Between 5th Maxima and 3rd Minima:**
- The separation \(p\) between the 5th maxima and the 3rd minima is given by:
\[ p = y_5 - y_3 \]
- Substituting the values we calculated:
\[ p = \frac{5 \lambda D}{d} - \frac{2.5 \lambda D}{d} \]
- Simplifying this gives:
\[ p = \frac{(5 - 2.5) \lambda D}{d} = \frac{2.5 \lambda D}{d} \]

5. **Calculating the Fringe Width (β):**
- The fringe width \(\beta\) is defined as:
\[ \beta = \frac{\lambda D}{d} \]

6. **Relating the Separation to Fringe Width:**
- Now, we can express \(p\) in terms of \(\beta\):
\[ p = 2.5 \cdot \frac{\lambda D}{d} = 2.5 \beta \]

7. **Final Answer:**
- Therefore, the separation between the 5th maxima and the 3rd minima is:
\[ p = 2.5 \beta \]
- This means the separation is **2.5 times the fringe width**.