The distance between two coherent sources is 0.1 mm. The fringe-width on a screen 1.2 m away from the source is 6.0 mm. The wavelength of light used is

Options

A. `4000 Å`
B. `5000 Å`
C. `6000 Å`
D. `7200 Å`

Explanation

To find the wavelength of light used in the given problem, we can follow these steps:

### Step 1: Write down the given values
- Distance between the two coherent sources (small d):
\[ d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \]
- Distance from the sources to the screen (capital D):
\[ D = 1.2 \, \text{m} \]
- Fringe width (beta):
\[ \beta = 6.0 \, \text{mm} = 6.0 \times 10^{-3} \, \text{m} \]

### Step 2: Use the formula for fringe width
The formula relating fringe width (\(\beta\)), wavelength (\(\lambda\)), distance between sources (\(d\)), and distance to the screen (\(D\)) is given by:
\[ \beta = \frac{d \lambda}{D} \]

### Step 3: Rearrange the formula to solve for wavelength (\(\lambda\))
Rearranging the formula gives:
\[ \lambda = \frac{\beta D}{d} \]

### Step 4: Substitute the known values into the formula
Substituting the values we have:
\[ \lambda = \frac{(6.0 \times 10^{-3} \, \text{m})(1.2 \, \text{m})}{1 \times 10^{-4} \, \text{m}} \]

### Step 5: Calculate the wavelength
Calculating the above expression:
\[ \lambda = \frac{7.2 \times 10^{-3}}{1 \times 10^{-4}} = 7.2 \times 10^{1} \, \text{m} = 720 \, \text{nm} \]

### Step 6: Convert to appropriate units
Since the wavelength is often expressed in nanometers (nm), we have:
\[ \lambda = 720 \, \text{nm} \]

### Final Result
The wavelength of light used is:
\[ \lambda = 720 \, \text{nm} \]