In a Young's double-slit experiment the fringe width is `0.2mm`. If the wavelength of light used is increased by 10% and the separation between the slits if also increased by 10%, the fringe width will be

Options

A. (a) `0.20mm`
B. (b) `0.401mm`
C. (c) `0.242mm`
D. (d) `0.165mm`

Explanation

To solve the problem, we will use the formula for fringe width in a Young's double-slit experiment, which is given by:

\[ \beta = \frac{D \lambda}{d} \]

Where:
- \(\beta\) is the fringe width,
- \(D\) is the distance from the slits to the screen,
- \(\lambda\) is the wavelength of the light,
- \(d\) is the separation between the slits.

### Step 1: Understand the initial conditions
The initial fringe width is given as:
\[ \beta_1 = 0.2 \, \text{mm} \]

### Step 2: Determine the changes in wavelength and slit separation
The problem states that both the wavelength (\(\lambda\)) and the slit separation (\(d\)) are increased by 10%.

- New wavelength:
\[ \lambda_2 = \lambda_1 \times 1.1 \]
- New slit separation:
\[ d_2 = d_1 \times 1.1 \]

### Step 3: Substitute the new values into the fringe width formula
The new fringe width (\(\beta_2\)) can be calculated as:
\[ \beta_2 = \frac{D \lambda_2}{d_2} \]

Substituting the new values:
\[ \beta_2 = \frac{D (1.1 \lambda_1)}{1.1 d_1} \]

### Step 4: Simplify the expression
Notice that the \(1.1\) in the numerator and denominator cancels out:
\[ \beta_2 = \frac{D \lambda_1}{d_1} = \beta_1 \]

### Step 5: Conclusion
Since \(\beta_2 = \beta_1\), the new fringe width remains the same:
\[ \beta_2 = 0.2 \, \text{mm} \]

### Final Answer:
The fringe width will be \(0.2 \, \text{mm}\).

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