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 will follow these 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 (µm). The introduction of this film causes the central fringe in a Young's Double Slit Experiment (YDSE) to move to the position of the 10th bright fringe.

### Step 2: Convert Thickness to Meters
Convert the thickness from micrometers to meters:
\[ T = 0.01 \, \mu m = 0.01 \times 10^{-6} \, m = 1 \times 10^{-8} \, m \]

### Step 3: Use the Path Difference Formula
The path difference introduced by the thin film is given by:
\[ \Delta y = (µ - 1) \cdot T \]
Substituting the values:
\[ \Delta y = (1.6 - 1) \cdot (1 \times 10^{-8}) = 0.6 \cdot (1 \times 10^{-8}) = 6 \times 10^{-9} \, m \]

### Step 4: Relate Path Difference to Fringe Movement
The movement of the central fringe to the position of the 10th bright fringe corresponds to a path difference of:
\[ \Delta y = 10 \cdot \lambda \]

### Step 5: Set Up the Equation
Equating the two expressions for path difference:
\[ 6 \times 10^{-9} = 10 \cdot \lambda \]

### Step 6: Solve for Wavelength (λ)
Rearranging the equation to find λ:
\[ \lambda = \frac{6 \times 10^{-9}}{10} = 6 \times 10^{-10} \, m \]

### Step 7: Convert Wavelength to Angstroms
Since 1 Angstrom (Å) = \(10^{-10}\) m, we convert the wavelength:
\[ \lambda = 6 \times 10^{-10} \, m = 6 \, \text{Å} \]

### Step 8: Conclusion
The wavelength of the wave is:
\[ \lambda = 6 \, \text{Å} \]

### Final Answer
The correct option is **6 Å**.

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