In YDSE a thin film `(mu=1.6)` of thickness`0.01 mu m` is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the 10th bright fring earlier. The wave length of wave is

Options

A. 6A
B. 6000A
C. 60A
D. 660A

Answer

To solve the problem, we need to find the wavelength of the light used in the Young's Double Slit Experiment (YDSE) after a thin film is introduced. Let's break down the steps:

### Step 1: Understand the Problem
We have a thin film with a refractive index (μ) of 1.6 and a thickness (t) of 0.01 micrometers (or 0.01 × 10^-6 meters). The introduction of this film causes the central fringe to shift to the position of the 10th bright fringe.

### Step 2: Calculate the Optical Path Difference
The optical path difference (OPD) introduced by the thin film can be calculated using the formula:
\[ \text{OPD} = (μ - 1) \cdot t \]
Substituting the values:
- μ = 1.6
- t = 0.01 × 10^-6 m

Calculating:
\[ \text{OPD} = (1.6 - 1) \cdot (0.01 \times 10^{-6}) \]
\[ \text{OPD} = 0.6 \cdot (0.01 \times 10^{-6}) \]
\[ \text{OPD} = 0.006 \times 10^{-6} \text{ m} \]
\[ \text{OPD} = 6 \times 10^{-9} \text{ m} \]

### Step 3: Relate the OPD to the Fringe Shift
The shift of the central fringe to the position of the 10th bright fringe indicates that the OPD is equal to 10 wavelengths (λ):
\[ \text{OPD} = 10 \lambda \]

### Step 4: Set Up the Equation
From the above relationship, we can set up the equation:
\[ 10 \lambda = 6 \times 10^{-9} \text{ m} \]

### Step 5: Solve for Wavelength (λ)
Now, we can solve for λ:
\[ \lambda = \frac{6 \times 10^{-9}}{10} \]
\[ \lambda = 0.6 \times 10^{-9} \text{ m} \]
\[ \lambda = 6 \times 10^{-10} \text{ m} \]
Converting this to Angstroms (1 Angstrom = 10^-10 m):
\[ \lambda = 6 \text{ Angstroms} \]

### Final Answer
The wavelength of the wave is **6 Angstroms**.

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