A light of wavelength `6000 Å` shines on two narrow slits separeted by a distance 1.0 mm and illuminates a screen at a distance 1.5 m away. When one slit is covered by a thin glass plate of refractive index 1.8 and other slit by a thin glass plate of refractive index `mu`, the central maxima shifts by 0.1 rad. Both plates have the same thickness of 0.5 mm. The value of refractive index `mu` of the glass is

Options

A. 1.4
B. 1.5
C. 1.6
D. none of these

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Understand the Problem
We have two slits illuminated by light of wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \). The distance between the slits is \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \) and the distance to the screen is \( D = 1.5 \, \text{m} \). One slit is covered with a glass plate of refractive index \( \mu_1 = 1.8 \) and the other with a glass plate of refractive index \( \mu_2 \). The thickness of both plates is \( t = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \). The central maximum shifts by an angle of \( \theta = 0.1 \, \text{rad} \).

### Step 2: Calculate the Shift in Position
The shift in position \( \Delta x \) on the screen due to the angle \( \theta \) can be calculated using the formula:
\[ \Delta x = D \tan(\theta) \]
For small angles, \( \tan(\theta) \approx \theta \), so:
\[ \Delta x \approx D \theta = 1.5 \times 0.1 = 0.15 \, \text{m} \]

### Step 3: Calculate the Optical Path Difference
The optical path difference caused by the glass plates can be expressed as:
\[ \Delta x = (\mu_2 - \mu_1) t \frac{D}{d} \]
Where \( \mu_1 = 1.8 \) and \( \mu_2 \) is the unknown refractive index.

### Step 4: Set Up the Equation
From the previous steps, we can equate the shifts:
\[ 0.15 = (\mu_2 - 1.8) \left(0.5 \times 10^{-3}\right) \frac{1.5}{1.0 \times 10^{-3}} \]

### Step 5: Simplify the Equation
Now, simplifying the equation:
\[ 0.15 = (\mu_2 - 1.8) \left(0.5 \times 10^{-3}\right) \cdot 1500 \]
\[ 0.15 = (\mu_2 - 1.8) \cdot 0.75 \]
\[ \mu_2 - 1.8 = \frac{0.15}{0.75} = 0.2 \]
\[ \mu_2 = 1.8 + 0.2 = 2.0 \]

### Step 6: Check for Validity
Since the problem states that the central maxima shifts by \( 0.1 \, \text{rad} \), we need to check if this value of \( \mu_2 \) is valid.

### Step 7: Final Calculation
If we take the negative shift (as the shift could be in either direction):
\[ \mu_2 - 1.8 = -0.2 \implies \mu_2 = 1.8 - 0.2 = 1.6 \]

### Conclusion
Thus, the value of the refractive index \( \mu \) is:
\[ \mu = 1.6 \]