In a certain double slit experimental arrangement interference fringes of width `1.0mm` each are observed when light of wavelength `5000Å` is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength `6000Å`, the fringe width will be

Options

A. 0.5 mm
B. 1.0 mm
C. 1.2 mm
D. 1.5 mm

Answer

To solve the problem, we need to use the formula for fringe width in a double slit experiment, which is given by:

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

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

### Step 1: Identify the initial fringe width and wavelength
From the problem, we know that the initial fringe width (\(\beta_1\)) is \(1.0 \, \text{mm}\) and the initial wavelength (\(\lambda_1\)) is \(5000 \, \text{Å}\).

### Step 2: Convert the initial wavelength to meters
We need to convert the wavelength from angstroms to meters for consistency in units:
\[ 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

### Step 3: Determine the new wavelength
The new wavelength (\(\lambda_2\)) is given as \(6000 \, \text{Å}\). We convert this to meters as well:
\[ 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

### Step 4: Use the relationship between fringe widths and wavelengths
Since the setup remains unchanged, the ratio of the fringe widths is directly proportional to the ratio of the wavelengths:
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \]

### Step 5: Substitute the known values
We know:
- \(\beta_1 = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}\)
- \(\lambda_1 = 5 \times 10^{-7} \, \text{m}\)
- \(\lambda_2 = 6 \times 10^{-7} \, \text{m}\)

Substituting these values into the ratio:
\[ \frac{\beta_2}{1.0 \times 10^{-3}} = \frac{6 \times 10^{-7}}{5 \times 10^{-7}} \]

### Step 6: Calculate the new fringe width
Now, we can solve for \(\beta_2\):
\[ \beta_2 = 1.0 \times 10^{-3} \times \frac{6}{5} = 1.0 \times 10^{-3} \times 1.2 = 1.2 \times 10^{-3} \, \text{m} = 1.2 \, \text{mm} \]

### Final Answer
The new fringe width when the source is replaced by another source of wavelength \(6000 \, \text{Å}\) will be \(1.2 \, \text{mm}\).

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