A thin mica sheet of thickness `2xx10^-6m` and refractive index `(mu=1.5)` is introduced in the path of the first wave. The wavelength of the wave used is `5000Å`. The central bright maximum will shift

Options

A. 2 fringes upward
B. 2 fringes downward
C. 10 fringes upwards
D. none of these

Answer

To determine the shift in the central bright maximum when a thin mica sheet is introduced, we can follow these steps:

### Step 1: Identify the given values
- Thickness of the mica sheet (t) = \(2 \times 10^{-6} \, \text{m}\)
- Refractive index of mica (\(\mu\)) = 1.5
- Wavelength of the wave (\(\lambda\)) = 5000 Å = \(5000 \times 10^{-10} \, \text{m}\)

### Step 2: Calculate the extra path difference
The extra path difference (\(\Delta x\)) introduced by the mica sheet can be calculated using the formula:
\[ \Delta x = (\mu - 1) \cdot t \]
Substituting the values:
\[ \Delta x = (1.5 - 1) \cdot (2 \times 10^{-6}) = 0.5 \cdot (2 \times 10^{-6}) = 1 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the number of fringes shifted
The number of fringes (\(n\)) shifted can be calculated using the formula:
\[ n = \frac{\Delta x}{\lambda} \]
Substituting the values:
\[ n = \frac{1 \times 10^{-6}}{5000 \times 10^{-10}} = \frac{1 \times 10^{-6}}{5 \times 10^{-7}} = 2 \]

### Step 4: Interpret the result
Since \(n = 2\), this means that the central bright maximum will shift by 2 fringes upwards.

### Final Answer
The central bright maximum will shift by 2 fringes upwards.

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