In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is `0.20 Wm^-2`, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Separation between the slits, \( d = 2.0 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Wavelength of the light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Distance of the screen from the slits, \( D = 2.0 \, \text{m} \)
- Intensity at the center of the central maximum, \( I_{\text{max}} = 0.20 \, \text{W/m}^2 \)
- Distance from the center to the point of interest, \( y = 0.5 \, \text{cm} = 0.5 \times 10^{-2} \, \text{m} \)

### Step 2: Calculate the path difference
The path difference \( \Delta \) at a distance \( y \) from the center can be calculated using the formula:

\[ \Delta = \frac{y \cdot d}{D} \]

Substituting the values:

\[ \Delta = \frac{(0.5 \times 10^{-2}) \cdot (2 \times 10^{-3})}{2} = \frac{1 \times 10^{-5}}{2} = 5 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the phase difference
The phase difference \( \phi \) can be calculated using the formula:

\[ \phi = \frac{2\pi}{\lambda} \cdot \Delta \]

Substituting the values:

\[ \phi = \frac{2\pi}{600 \times 10^{-9}} \cdot (5 \times 10^{-6}) = \frac{10\pi}{600} = \frac{\pi}{60} \, \text{radians} \]

### Step 4: Calculate the resultant amplitude
The resultant amplitude \( A \) can be calculated using the formula:

\[ A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi)} \]

Assuming \( A_1 = A_2 = A \) (the amplitude from each slit is the same):

\[ A = \sqrt{A^2 + A^2 + 2A^2 \cos(\phi)} = \sqrt{2A^2(1 + \cos(\phi))} = A\sqrt{2(1 + \cos(\phi))} \]

Using \( \cos(\phi) = \cos\left(\frac{\pi}{60}\right) \):

\[ A = A\sqrt{2(1 + \cos\left(\frac{\pi}{60}\right))} \]

### Step 5: Calculate the intensity at the point
The intensity \( I \) at the point can be calculated using the ratio of the resultant amplitude to the maximum amplitude:

\[ \frac{I}{I_{\text{max}}} = \left(\frac{A}{2A}\right)^2 = \frac{1}{4} \]

Thus,

\[ I = \frac{I_{\text{max}}}{4} = \frac{0.20}{4} = 0.05 \, \text{W/m}^2 \]

### Final Answer
The intensity at the point 0.5 cm away from the center along the width of the fringes is:

\[ \boxed{0.05 \, \text{W/m}^2} \]