YDSE is conducted using light of wavelength `6000 Å` to observe an interference pattern. When a film of some material `3.0xx10^(-3)` cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 10 fringe widths. What is the refractive index of the material of the film ?

Answer

To solve the problem, we need to determine the refractive index of the material of the film based on the given information about the interference pattern in Young's Double Slit Experiment (YDSE).

### Given:
- Wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Thickness of the film, \( t = 3.0 \times 10^{-3} \, \text{cm} = 3.0 \times 10^{-5} \, \text{m} \)
- Fringe shift, \( \Delta x = 10 \, \text{fringe widths} \)

### Step 1: Calculate the fringe width
The fringe width \( \beta \) in YDSE is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. However, we do not need to know \( D \) and \( d \) explicitly, as we will relate the shift to the fringe width.

### Step 2: Calculate the total optical path difference caused by the film
When a film of refractive index \( n \) is placed over one of the slits, the optical path difference (OPD) introduced by the film is given by:
\[ \text{OPD} = (n - 1) t \]

### Step 3: Relate the shift in fringe pattern to the optical path difference
The shift in the fringe pattern \( \Delta x \) is related to the optical path difference by:
\[ \Delta x = \frac{\text{OPD}}{\beta} \]
Given that the fringe pattern shifted by a distance equal to 10 fringe widths, we can write:
\[ \Delta x = 10 \beta \]

### Step 4: Substitute the expressions into the equation
Now substituting the expressions for \( \Delta x \) and OPD:
\[ 10 \beta = \frac{(n - 1) t}{\beta} \]
Rearranging gives:
\[ 10 \beta^2 = (n - 1) t \]

### Step 5: Solve for the refractive index \( n \)
Now, we can express \( n \):
\[ n - 1 = \frac{10 \beta^2}{t} \]
\[ n = 1 + \frac{10 \beta^2}{t} \]

### Step 6: Calculate \( \beta \)
From the earlier relationship, we know:
\[ \beta = \frac{\lambda D}{d} \]
Since we do not have \( D \) and \( d \), we can express \( \beta^2 \) in terms of \( \lambda \):
\[ \beta^2 = \left(\frac{\lambda D}{d}\right)^2 \]
However, we can also express \( \beta \) in terms of the fringe shift:
\[ \Delta x = 10 \beta \implies \beta = \frac{\Delta x}{10} \]
Substituting \( \Delta x = 10 \beta \) gives:
\[ \beta = \frac{(n - 1)t}{10} \]

### Step 7: Substitute \( \beta \) back into the equation for \( n \)
Substituting \( \beta \) back into the equation for \( n \):
\[ n = 1 + \frac{10 \left(\frac{\Delta x}{10}\right)^2}{t} \]
Substituting \( \Delta x = 10 \beta \):
\[ n = 1 + \frac{10 \left(\frac{10 \beta}{10}\right)^2}{t} \]

### Final Calculation
Now substituting \( t = 3.0 \times 10^{-5} \, \text{m} \) and \( \lambda = 6.0 \times 10^{-7} \, \text{m} \):
\[ \beta = \frac{6.0 \times 10^{-7}}{10} = 6.0 \times 10^{-8} \, \text{m} \]
\[ n = 1 + \frac{10 \cdot (6.0 \times 10^{-8})^2}{3.0 \times 10^{-5}} \]
Calculating \( n \):
\[ n = 1 + \frac{10 \cdot 36 \times 10^{-16}}{3.0 \times 10^{-5}} = 1 + \frac{360 \times 10^{-16}}{3.0 \times 10^{-5}} = 1 + 1.2 = 2.2 \]

### Conclusion
The refractive index of the material of the film is approximately \( n = 2.2 \).