In YDSE, both slits produce equal intensities on the screen. A 100% transparent thin film is placed in front of one of the slits. Now, the intensity on the centre becomes 75% of the previous intensity. The wavelength of light is 6000Å and refractive index of glass is 1.5. Thus, minimum thickness of the glass slab is

Options

A. `0.2mum`
B. `0.3mum`
C. `0.4mum`
D. `0.5 mu m`

Explanation

To solve the problem, we need to find the minimum thickness of the glass slab that causes the intensity at the center of the interference pattern to reduce to 75% of its original value. Here’s the step-by-step solution:

### Step 1: Understand the Initial Condition
In the Young's Double Slit Experiment (YDSE), both slits produce equal intensities on the screen. Let's denote the initial intensity at the center as \( I_0 \).

### Step 2: Determine the New Intensity
After placing the 100% transparent thin film in front of one of the slits, the intensity at the center becomes 75% of the original intensity. Therefore, the new intensity \( I \) can be expressed as:
\[ I = 0.75 I_0 = \frac{3}{4} I_0 \]

### Step 3: Relate Intensity to Phase Difference
In YDSE, the intensity at the center can be expressed in terms of the phase difference \( \phi \) as:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_{\text{max}} \) is the maximum intensity.

### Step 4: Set Up the Equation
Since the new intensity is \( \frac{3}{4} I_{\text{max}} \), we can set up the equation:
\[ \frac{3}{4} I_{\text{max}} = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_{\text{max}} \) gives:
\[ \frac{3}{4} = \cos^2\left(\frac{\phi}{2}\right) \]

### Step 5: Solve for Phase Difference
Taking the square root of both sides, we find:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{6} \quad \Rightarrow \quad \phi = \frac{\pi}{3} \]

### Step 6: Relate Phase Difference to Path Difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the equation:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{\pi}{3} \):
\[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{6} \]

### Step 7: Calculate the Path Difference Due to the Glass Slab
When a glass slab of thickness \( t \) is placed in front of one of the slits, the effective path difference \( \Delta x \) is given by:
\[ \Delta x = (n - 1) t \]
where \( n \) is the refractive index of the glass. Here, \( n = 1.5 \).

### Step 8: Set Up the Equation for Thickness
Equating the two expressions for path difference:
\[ (n - 1) t = \frac{\lambda}{6} \]
Substituting \( n = 1.5 \) and \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \):
\[ (1.5 - 1) t = \frac{6000 \times 10^{-10}}{6} \]
\[ 0.5 t = \frac{6000 \times 10^{-10}}{6} \]
\[ t = \frac{6000 \times 10^{-10}}{3} = 2000 \times 10^{-10} \, \text{m} = 2000 \, \text{Å} \]

### Final Answer
The minimum thickness of the glass slab is:
\[ \boxed{2000 \, \text{Å}} \quad \text{or} \quad 0.2 \, \mu m \]