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

### Step 1: Understand the formula for dark fringes
The position of the dark fringes in an interference pattern is given by the formula:
\[ y_n = \frac{(2n + 1) \lambda D}{2d} \]
where:
- \(y_n\) is the distance of the nth dark fringe from the center,
- \(n\) is the order of the dark fringe (for the 5th dark fringe, \(n = 5\)),
- \(\lambda\) is the wavelength of light,
- \(D\) is the distance from the slits to the screen,
- \(d\) is the distance between the slits.

### Step 2: Substitute for the 5th dark fringe
For the 5th dark fringe, we substitute \(n = 5\):
\[ y_5 = \frac{(2 \cdot 5 + 1) \lambda D}{2d} = \frac{11 \lambda D}{2d} \]

### Step 3: Relate fringe width to the parameters
The fringe width \(\beta\) is given by:
\[ \beta = \frac{\lambda D}{d} \]
This means we can express \(\lambda D\) in terms of \(\beta\) and \(d\):
\[ \lambda D = \beta \cdot d \]

### Step 4: Substitute \(\lambda D\) into the dark fringe formula
Now, substituting \(\lambda D\) into the equation for \(y_5\):
\[ y_5 = \frac{11 (\beta \cdot d)}{2d} = \frac{11 \beta}{2} \]

### Step 5: Calculate the distance for the given fringe width
Given that \(\beta = 0.002\) cm, we can now calculate \(y_5\):
\[ y_5 = \frac{11 \cdot 0.002 \text{ cm}}{2} = \frac{0.022 \text{ cm}}{2} = 0.011 \text{ cm} \]

### Final Answer
The distance of the 5th dark fringe from the center is:
\[ y_5 = 0.011 \text{ cm} \]

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