In a YDSE, fringes are produced by monochromatic light of wavelength 5450Å. A thin plate of glass of refractive index 1.5 is placed normally in the path of one of the intefering beams and the central bright band of the fringe system is found to move into the position previoysly occupied by the third band from the centre. select the correct alternatie

Options

A. The thickness of the plate is `3.27xx10^(-6)m`
B. If the separation between the sources is `1000Å` , the angular position of first maxima is 0.01 rad
C. The thickness of the plate is `1.59xx10^(-6)m`
D. A maxima is still obtained at the centre of screen

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
In a Young's double slit experiment (YDSE), we have two coherent light sources that create an interference pattern on a screen. The central bright fringe is where the path difference between the two beams is zero.

### Step 2: Identify the effect of the glass plate
When a thin glass plate of refractive index \( \mu = 1.5 \) is placed in the path of one of the beams, it introduces an additional optical path length. The effective path difference introduced by the glass plate is given by:

\[ \Delta g = t(\mu - 1) \]

where \( t \) is the thickness of the glass plate.

### Step 3: Determine the shift in fringe position
The problem states that the central bright band moves to the position of the third band. This means that the original central maximum (n=0) now coincides with the position of the third maximum (n=3). The path difference for the new position of the central maximum can be expressed as:

\[ \Delta = \Delta g - n\lambda = 0 \]

Substituting for \( \Delta g \):

\[ t(\mu - 1) - 3\lambda = 0 \]

### Step 4: Solve for thickness \( t \)
Rearranging the equation gives:

\[ t(\mu - 1) = 3\lambda \]

Now, substituting the values:

- Wavelength \( \lambda = 5450 \, \text{Å} = 5450 \times 10^{-10} \, \text{m} \)
- Refractive index \( \mu = 1.5 \)

We can calculate \( t \):

\[ t = \frac{3\lambda}{\mu - 1} = \frac{3 \times 5450 \times 10^{-10}}{1.5 - 1} \]

Calculating \( \mu - 1 \):

\[ \mu - 1 = 0.5 \]

Now substituting this back into the equation for \( t \):

\[ t = \frac{3 \times 5450 \times 10^{-10}}{0.5} = 3 \times 3.27 \times 10^{-6} \, \text{m} = 9.81 \times 10^{-6} \, \text{m} \]

### Step 5: Conclusion
The thickness of the glass plate is:

\[ t = 3.27 \times 10^{-6} \, \text{m} \]

### Step 6: Analyze the options
1. **Option A**: The thickness of the plate is \( 3.27 \times 10^{-6} \, \text{m} \) (Correct)
2. **Option B**: The thickness of the plate is \( 1.59 \times 10^{-6} \, \text{m} \) (Incorrect)
3. **Option C**: Not provided in the transcript.
4. **Option D**: A maxima is still obtained at the center of the screen (Correct)

### Final Answer
The correct alternatives are **Option A** and **Option D**.

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