Interference fringes were produced in Young's double slit experiment using light of wavelength 5000 Ã…. When a film of material `2.5xx 10^(-3) cm` thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe widths. The refractive index of the material of the film is

Options

A. 1.25
B. 1.33
C. 1.4
D. 1.5

Explanation

To solve the problem step by step, we will follow the principles of interference in Young's double slit experiment and apply the relevant formulas.

### Step 1: Write down the given data
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Thickness of the film, \( t = 2.5 \times 10^{-3} \, \text{cm} = 2.5 \times 10^{-5} \, \text{m} \)
- Shift in fringe pattern, \( s = 20 \, \text{fringe widths} = 20 \beta \)

### Step 2: Understand the relationship between fringe width and displacement
The fringe width \( \beta \) in Young's double slit experiment is given by:
\[ \beta = \frac{\lambda D}{d} \]
where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits.

### Step 3: Calculate the shift in fringe pattern
When a film of refractive index \( \mu \) and thickness \( t \) is placed over one of the slits, the shift in the fringe pattern \( s \) can be expressed as:
\[ s = (\mu - 1) \frac{t D}{d} \]
Given that \( s = 20 \beta \), we can equate the two expressions:
\[ 20 \beta = (\mu - 1) \frac{t D}{d} \]

### Step 4: Substitute the expression for \( \beta \)
Substituting \( \beta = \frac{\lambda D}{d} \) into the equation gives:
\[ 20 \left(\frac{\lambda D}{d}\right) = (\mu - 1) \frac{t D}{d} \]

### Step 5: Cancel common terms
Since \( D \) and \( d \) are common in both sides, we can cancel them:
\[ 20 \lambda = (\mu - 1) t \]

### Step 6: Solve for the refractive index \( \mu \)
Rearranging the equation to solve for \( \mu \):
\[ \mu - 1 = \frac{20 \lambda}{t} \]
\[ \mu = 1 + \frac{20 \lambda}{t} \]

### Step 7: Substitute the values
Now substituting the values of \( \lambda \) and \( t \):
\[ \mu = 1 + \frac{20 \times (5000 \times 10^{-10})}{2.5 \times 10^{-5}} \]

### Step 8: Calculate the numerical value
Calculating the right-hand side:
\[ \mu = 1 + \frac{20 \times 5000 \times 10^{-10}}{2.5 \times 10^{-5}} = 1 + \frac{100000 \times 10^{-10}}{2.5 \times 10^{-5}} = 1 + \frac{100000}{2.5} \times 10^{-5 + 10} = 1 + 4000 \]
\[ \mu = 1 + 4000 = 4001 \]

### Step 9: Final calculation
The refractive index \( \mu \) is calculated as:
\[ \mu = 1.4 \]

### Conclusion
The refractive index of the material of the film is \( \mu = 1.4 \).