Two coherent light sources `S_1` and `S_2 (lambda=6000Å)` are `1mm` apart from each other. The screen is placed at a distance of `25cm` from the sources. The width of the fringes on the screen should be

Options

A. 0.015 cm
B. 0.025 cm
C. 0.010 cm
D. 0.030 cm

Explanation

To find the width of the fringes on the screen, we can use the formula for fringe width (β) in a double-slit interference pattern:

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

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

### Step 1: Identify the given values
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6000 \times 10^{-8} \, \text{cm} = 6 \times 10^{-5} \, \text{cm} \)
- Distance between the slits \( d = 1 \, \text{mm} = 0.1 \, \text{cm} \)
- Distance to the screen \( D = 25 \, \text{cm} \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the formula:

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

Substituting the known values:

\[ \beta = \frac{25 \, \text{cm} \cdot 6 \times 10^{-5} \, \text{cm}}{0.1 \, \text{cm}} \]

### Step 3: Calculate the fringe width
Calculating the numerator:

\[ 25 \cdot 6 = 150 \]

So we have:

\[ \beta = \frac{150 \times 10^{-5} \, \text{cm}}{0.1 \, \text{cm}} \]

Now, dividing by \( 0.1 \):

\[ \beta = 1500 \times 10^{-5} \, \text{cm} = 1.5 \times 10^{-2} \, \text{cm} \]

### Step 4: Convert to a more convenient unit
To express this in centimeters:

\[ \beta = 0.015 \, \text{cm} \]

### Final Answer
The width of the fringes on the screen is \( 0.015 \, \text{cm} \).

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