In Young's double slit experiment, the fringes are displaced index 1.5 is introduced in the path of one of the beams. When this plate in replaced by another plate of the same thickness, the shift of fringes is `(3//2)x`. The refractive index of the second plate is

Options

A. 1.75
B. `1.40`
C. `1.25`
D. 1.67

Answer

To solve the problem, we need to analyze the situation in Young's double slit experiment when a plate with a refractive index is introduced in the path of one of the beams. Let's break down the solution step by step.

### Step 1: Understand the fringe shift formula
In Young's double slit experiment, the fringe shift (x) caused by introducing a plate of refractive index (μ) in the path of one of the beams is given by the formula:

\[ x = (μ - 1) \frac{T}{\lambda} \]

where:
- \( μ \) is the refractive index of the plate,
- \( T \) is the thickness of the plate,
- \( \lambda \) is the wavelength of light used.

### Step 2: Apply the formula for the first plate
For the first plate with a refractive index \( μ_1 = 1.5 \), the fringe shift \( x_1 \) can be expressed as:

\[ x_1 = (1.5 - 1) \frac{T}{\lambda} = 0.5 \frac{T}{\lambda} \]

### Step 3: Apply the formula for the second plate
When the first plate is replaced by a second plate of the same thickness \( T \) but with a different refractive index \( μ_2 \), the fringe shift \( x_2 \) is given as:

\[ x_2 = (μ_2 - 1) \frac{T}{\lambda} \]

According to the problem, the shift of fringes with the second plate is \( \frac{3}{2} x_1 \):

\[ x_2 = \frac{3}{2} x_1 \]

### Step 4: Substitute the expression for \( x_1 \) into \( x_2 \)
Substituting the expression for \( x_1 \) into the equation for \( x_2 \):

\[ (μ_2 - 1) \frac{T}{\lambda} = \frac{3}{2} \left(0.5 \frac{T}{\lambda}\right) \]

### Step 5: Simplify the equation
This simplifies to:

\[ (μ_2 - 1) \frac{T}{\lambda} = \frac{3}{4} \frac{T}{\lambda} \]

### Step 6: Cancel out common terms
Since \( \frac{T}{\lambda} \) is common on both sides, we can cancel it out:

\[ μ_2 - 1 = \frac{3}{4} \]

### Step 7: Solve for \( μ_2 \)
Now, solving for \( μ_2 \):

\[ μ_2 = 1 + \frac{3}{4} = \frac{4}{4} + \frac{3}{4} = \frac{7}{4} = 1.75 \]

### Final Answer
The refractive index of the second plate \( μ_2 \) is:

\[ \boxed{1.75} \]