Interference fringes were produced in Young's double-slit experiment using light of wavelength `5000 Å`. When a film of thickness `2.5xx10^(-3) cm` was placed in front of 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. `1.35`
C. `1.4`
D. `1.5`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
- Thickness of the film, \( t = 2.5 \times 10^{-3} \, \text{cm} = 2.5 \times 10^{-5} \, \text{m} \)
- The fringe pattern shifts by \( 20 \) fringe-widths.

### Step 2: Relate the shift in fringe pattern to the refractive index
The shift in the fringe pattern due to the introduction of the film is given by the formula:
\[ \Delta y = (n - 1) \cdot t \cdot \frac{D}{d} \]
where:
- \( \Delta y \) is the shift in the fringe pattern,
- \( n \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 3: Express the fringe width
The fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \]
Thus, the shift \( \Delta y \) can also be expressed in terms of fringe width:
\[ \Delta y = 20 \cdot \beta = 20 \cdot \frac{\lambda D}{d} \]

### Step 4: Set the two expressions for shift equal
Equating the two expressions for the shift, we have:
\[ (n - 1) \cdot t \cdot \frac{D}{d} = 20 \cdot \frac{\lambda D}{d} \]

### Step 5: Cancel common terms
Since \( \frac{D}{d} \) appears on both sides, we can cancel it out:
\[ (n - 1) \cdot t = 20 \cdot \lambda \]

### Step 6: Solve for the refractive index \( n \)
Rearranging the equation gives:
\[ n - 1 = \frac{20 \cdot \lambda}{t} \]
Thus,
\[ n = 1 + \frac{20 \cdot \lambda}{t} \]

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

### Step 8: Calculate the value
Calculating the fraction:
\[ \frac{20 \cdot (5 \times 10^{-7})}{2.5 \times 10^{-5}} = \frac{100 \times 10^{-7}}{2.5 \times 10^{-5}} = \frac{100}{2.5} \cdot 10^{-2} = 40 \cdot 10^{-2} = 0.4 \]
Thus,
\[ n = 1 + 0.4 = 1.4 \]

### Final Answer
The refractive index of the material of the film is \( n = 1.4 \).

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