When a thin transparent plate of Refractive Index 1.5 is introduced in one of the interfearing becomes, 20 fringes shift. If it is replaced by another thin plate of half the thickness and of R.I 1.7 the number of fringes that undergo displacement is

Options

A. 23
B. 14
C. 28
D. 7

Answer

To solve the problem step-by-step, we will use the formula for the number of fringes shifted in an interference pattern when a thin transparent plate is introduced.

### Step 1: Understand the formula for fringe shift
The number of fringes shifted (n) when a thin transparent plate of thickness \( t \) and refractive index \( \mu \) is introduced is given by:
\[ n = \frac{(\mu - 1) \cdot t}{\lambda} \]
where \( \lambda \) is the wavelength of the light used.

### Step 2: Calculate the number of fringes shifted for the first plate
For the first plate:
- Refractive index \( \mu_1 = 1.5 \)
- Let the thickness of the plate be \( t \).
- The number of fringes shifted is given as \( n_1 = 20 \).

Using the formula:
\[ n_1 = \frac{(1.5 - 1) \cdot t}{\lambda} = \frac{0.5 \cdot t}{\lambda} \]
We can express this as:
\[ 20 = \frac{0.5 \cdot t}{\lambda} \quad \text{(1)} \]

### Step 3: Set up the equation for the second plate
For the second plate:
- Refractive index \( \mu_2 = 1.7 \)
- The thickness of the second plate is half of the first, so \( t_2 = \frac{t}{2} \).

Using the same formula for the second plate:
\[ n_2 = \frac{(1.7 - 1) \cdot t_2}{\lambda} = \frac{(0.7) \cdot \frac{t}{2}}{\lambda} \]
This simplifies to:
\[ n_2 = \frac{0.7 \cdot t}{2\lambda} \quad \text{(2)} \]

### Step 4: Relate the two equations
From equation (1), we have:
\[ \frac{t}{\lambda} = 40 \quad \text{(from rearranging 20 = 0.5 * t / λ)} \]

Now substituting this into equation (2):
\[ n_2 = \frac{0.7 \cdot \left(40 \lambda\right)}{2\lambda} \]
The \( \lambda \) cancels out:
\[ n_2 = \frac{0.7 \cdot 40}{2} = \frac{28}{2} = 14 \]

### Conclusion
The number of fringes that undergo displacement when the second plate is introduced is approximately **14**.