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. 1
B. 2
C. 5
D. 10

Answer

To solve the problem step by step, we will follow the logical sequence of calculations based on the information provided in the question.

### 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, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)

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

### Step 3: Relate the extra path difference to the shift in the central maximum
The shift in the central maximum can be expressed in terms of the wavelength:
\[ \text{Shift} = n \lambda \]
where \( n \) is the number of wavelengths corresponding to the extra path difference.

### Step 4: Calculate \( n \)
From the previous step, we have:
\[ n = \frac{\text{Extra Path Difference}}{\lambda} \]
Substituting the values:
\[ n = \frac{1 \times 10^{-6}}{5000 \times 10^{-10}} = \frac{1 \times 10^{-6}}{5 \times 10^{-7}} = 2 \]

### Step 5: Conclusion
The central bright maximum will shift by \( n = 2 \) fringes.

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

---