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 step by step, we will follow the reasoning and calculations outlined in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a Young's double-slit experiment (YDSE) where both slits produce equal intensities on the screen.
- A 100% transparent thin film (glass slab) is placed in front of one of the slits.
- The intensity at the center of the screen becomes 75% of the previous maximum intensity.

2. **Setting Up the Intensity Equation:**
- Let the maximum intensity before placing the glass slab be \( I_{\text{max}} \).
- After placing the glass slab, the intensity at the center becomes \( \frac{3}{4} I_{\text{max}} \).
- The relationship between the intensities can be expressed as:
\[ \frac{3}{4} I_{\text{max}} = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
- Dividing both sides by \( I_{\text{max}} \) (assuming \( I_{\text{max}} \neq 0 \)):
\[ \frac{3}{4} = \cos^2\left(\frac{\phi}{2}\right) \]

3. **Finding the Phase Difference:**
- Taking the square root of both sides gives:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2} \]
- Therefore, we can find \( \frac{\phi}{2} \):
\[ \frac{\phi}{2} = \frac{\pi}{6} \quad \Rightarrow \quad \phi = \frac{\pi}{3} \]

4. **Calculating the Path Difference:**
- The path difference \( \Delta x \) corresponding to the phase difference \( \phi \) is given by:
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \phi \]
- Substituting \( \phi = \frac{\pi}{3} \):
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \frac{\pi}{3} = \frac{\lambda}{6} \]

5. **Path Difference Due to the Glass Slab:**
- The path difference introduced by the glass slab is given by:
\[ \Delta x = (n - 1) t \]
- Here, \( n \) is the refractive index of the glass (1.5), and \( t \) is the thickness of the glass slab.
- Therefore, we can set the two expressions for path difference equal:
\[ (n - 1) t = \frac{\lambda}{6} \]

6. **Substituting Values:**
- Given \( n = 1.5 \) and \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \):
\[ (1.5 - 1) t = \frac{6000 \times 10^{-10}}{6} \]
- Simplifying:
\[ 0.5 t = \frac{6000 \times 10^{-10}}{6} \]
\[ t = \frac{6000 \times 10^{-10}}{3} = 2000 \times 10^{-10} \, \text{m} = 2000 \, \text{Å} \]

7. **Final Answer:**
- The minimum thickness of the glass slab is:
\[ t = 2000 \, \text{Å} = 0.2 \, \mu m \]