A monochromatic light of `lambda = 5000 Å` is incident on two identical slits separated by a distance of `5 xx 10^(-4)` m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of slits. A thin glass plate of thickness `1.5 xx 10^(-6)` m and refractive index `mu = 1.5` is placed between one of the slits and the screen . Find the intensity at the center of the screen.

Explanation

To solve the problem step by step, we will follow the outlined process to find the intensity at the center of the screen when a thin glass plate is introduced in the path of one of the beams in a double-slit interference experiment.

### Step 1: Identify Given Values
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
- Distance between the slits, \( d = 5 \times 10^{-4} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)
- Thickness of the glass plate, \( t = 1.5 \times 10^{-6} \, \text{m} \)
- Refractive index of the glass plate, \( \mu = 1.5 \)

### Step 2: Calculate the Path Difference
When a glass plate is introduced in the path of one of the beams, the effective path difference \( \Delta x \) is given by:
\[ \Delta x = (S_2O - S_1O) + (\mu - 1) t \]
Here, \( S_2O \) is the path through the air, and \( S_1O \) is the path through the glass. Since \( S_1O \) does not pass through the glass, we can simplify this to:
\[ \Delta x = 0 + (\mu - 1) t = (1.5 - 1) \times 1.5 \times 10^{-6} \]
Calculating this gives:
\[ \Delta x = 0.5 \times 1.5 \times 10^{-6} = 0.75 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the Phase Difference
The phase difference \( \phi \) corresponding to the path difference is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the values:
\[ \phi = \frac{2\pi}{5 \times 10^{-7}} \times 0.75 \times 10^{-6} \]
Calculating this gives:
\[ \phi = \frac{2\pi \times 0.75}{5} = \frac{1.5\pi}{5} = 0.3\pi \, \text{radians} \]

### Step 4: Calculate the Resultant Intensity
The intensity at the center of the screen can be calculated using the formula:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
Since \( I_{\text{max}} \) for two coherent sources is typically \( 4I_0 \) (where \( I_0 \) is the intensity from one slit), we will use:
\[ I = 4I_0 \cos^2\left(\frac{0.3\pi}{2}\right) \]
Calculating \( \frac{0.3\pi}{2} = 0.15\pi \):
\[ \cos(0.15\pi) \approx \cos(0.4712) \approx 0.891 \]
Thus:
\[ I = 4I_0 (0.891)^2 \approx 4I_0 \times 0.792 \approx 3.168I_0 \]

### Step 5: Conclusion
The intensity at the center of the screen is:
\[ I \approx 3.168I_0 \]