A fringe width of a certain interference pattern is `beta=0.002` cm What is the distance of 5th dark fringe centre?

Options

A. `1xx10^(-2)cm`
B. `11xx10^(-2)cm`
C. `1.1xx10^(-2)cm`
D. `3.28xx10^(6)cm`

Explanation

To solve the problem of finding the distance of the 5th dark fringe from the center in an interference pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Fringe width (β) = 0.002 cm
- We need to find the distance of the 5th dark fringe (n = 5) from the center.

2. **Formula for the Distance of the nth Dark Fringe**:
- The distance of the nth dark fringe from the center is given by:
\[ y_n = \frac{(2n + 1) \lambda d}{2d} \]
- However, we can relate this to the fringe width (β) using the formula:
\[ \beta = \frac{\lambda d}{d} \]
- Thus, we can express the distance of the nth dark fringe in terms of β:
\[ y_n = \frac{(2n + 1)}{2} \beta \]

3. **Substituting for the 5th Dark Fringe**:
- For the 5th dark fringe, n = 5:
\[ y_5 = \frac{(2 \cdot 5 + 1)}{2} \beta \]
- Simplifying this gives:
\[ y_5 = \frac{(10 + 1)}{2} \beta = \frac{11}{2} \beta \]

4. **Substituting the Value of β**:
- Now substitute the value of β:
\[ y_5 = \frac{11}{2} \times 0.002 \text{ cm} \]

5. **Calculating the Distance**:
- Perform the multiplication:
\[ y_5 = \frac{11 \times 0.002}{2} = \frac{0.022}{2} = 0.011 \text{ cm} \]
- This can also be expressed in scientific notation:
\[ y_5 = 1.1 \times 10^{-2} \text{ cm} \]

6. **Final Answer**:
- The distance of the 5th dark fringe from the center is:
\[ y_5 = 0.011 \text{ cm} \text{ or } 1.1 \times 10^{-2} \text{ cm} \]