Two slits are separated by a distance of `0.5mm` and illuminated with light of `lambda=6000Å`. If the screen is placed `2.5m` from the slits. The distance of the third bright image from the centre will be

Options

A. (a) `1.5mm`
B. (b) `3mm`
C. (c) `6mm`
D. (d) `9mm`

Explanation

To solve the problem step by step, we need to use the formula for the position of bright fringes in a double-slit interference pattern. The formula for the position of the nth bright fringe (or image) from the center is given by:

\[ x_n = \frac{n \lambda D}{d} \]

where:
- \( x_n \) = distance of the nth bright fringe from the center
- \( n \) = order of the bright fringe (in this case, \( n = 3 \))
- \( \lambda \) = wavelength of the light (in meters)
- \( D \) = distance from the slits to the screen (in meters)
- \( d \) = distance between the slits (in meters)

### Step 1: Convert the given values to appropriate units
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Distance between the slits \( d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \)
- Distance from the slits to the screen \( D = 2.5 \, \text{m} \)

### Step 2: Substitute the values into the formula
Now, we can substitute the values into the formula for the third bright fringe (\( n = 3 \)):

\[ x_3 = \frac{3 \cdot (6 \times 10^{-7} \, \text{m}) \cdot (2.5 \, \text{m})}{5 \times 10^{-4} \, \text{m}} \]

### Step 3: Calculate \( x_3 \)
Calculating the numerator:

\[ 3 \cdot (6 \times 10^{-7}) \cdot (2.5) = 4.5 \times 10^{-6} \, \text{m} \]

Now, divide by \( d \):

\[ x_3 = \frac{4.5 \times 10^{-6}}{5 \times 10^{-4}} = 9 \times 10^{-3} \, \text{m} \]

### Step 4: Convert \( x_3 \) to millimeters
To convert meters to millimeters:

\[ x_3 = 9 \times 10^{-3} \, \text{m} = 9 \, \text{mm} \]

### Final Answer
The distance of the third bright image from the center is \( 9 \, \text{mm} \).

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