Two slits at a distance of `1mm` are illuminated by a light of wavelength `6.5xx10^-7m`. The interference fringes are observed on a screen placed at a distance of `1m`. The distance between third dark fringe and fifth bright fringe will be

Options

A. `0.65 mm`
B. `1.63 mm`
C. `3.25 mm`
D. `4.88 mm`

Explanation

To solve the problem of finding the distance between the third dark fringe and the fifth bright fringe in a double-slit interference pattern, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Distance between the slits, \( d = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Wavelength of light, \( \lambda = 6.5 \times 10^{-7} \text{ m} \)
- Distance from the slits to the screen, \( D = 1 \text{ m} \)

2. **Calculate the Fringe Width (β):**
The fringe width \( \beta \) is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the values:
\[ \beta = \frac{(6.5 \times 10^{-7} \text{ m})(1 \text{ m})}{1 \times 10^{-3} \text{ m}} = 6.5 \times 10^{-4} \text{ m} = 0.65 \text{ mm} \]

3. **Determine the Position of the Bright and Dark Fringes:**
The position of the \( n \)-th bright fringe is given by:
\[ y_n = n \beta \]
The position of the \( m \)-th dark fringe is given by:
\[ y_m = \left(m + \frac{1}{2}\right) \beta \]

4. **Calculate the Positions:**
- For the 5th bright fringe (\( n = 5 \)):
\[ y_5 = 5 \beta = 5 \times (6.5 \times 10^{-4} \text{ m}) = 3.25 \times 10^{-3} \text{ m} = 3.25 \text{ mm} \]
- For the 3rd dark fringe (\( m = 3 \)):
\[ y_3 = \left(3 + \frac{1}{2}\right) \beta = \left(3.5\right) \beta = 3.5 \times (6.5 \times 10^{-4} \text{ m}) = 2.275 \times 10^{-3} \text{ m} = 2.275 \text{ mm} \]

5. **Calculate the Distance Between the 3rd Dark Fringe and the 5th Bright Fringe:**
\[ \Delta y = y_5 - y_3 = 3.25 \text{ mm} - 2.275 \text{ mm} = 0.975 \text{ mm} \]

### Final Answer:
The distance between the third dark fringe and the fifth bright fringe is \( 0.975 \text{ mm} \).

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