Young's double slit experiment is made in a liquid. The tenth bright fringe in liquid lies in screen where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately

Options

A. 1.8
B. 1.5
C. 1.67
D. 1.2

Answer

To solve the problem, we need to find the refractive index of the liquid (µL) based on the information provided about the Young's double slit experiment.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- In the experiment, we have two media: liquid (with refractive index µL) and vacuum (with refractive index µ0 = 1).
- The 10th bright fringe in the liquid corresponds to the 6th dark fringe in vacuum.

2. **Formulas for Bright and Dark Fringes**:
- The position of the nth bright fringe in a medium is given by:
\[ y_b = \frac{n \lambda D}{\mu L} \]
- The position of the mth dark fringe in a medium is given by:
\[ y_d = \frac{(2m - 1) \lambda D}{2} \]
- Here, \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( n \) and \( m \) are the fringe orders.

3. **Setting Up the Equation**:
- For the 10th bright fringe in liquid:
\[ y_{10} = \frac{10 \lambda D}{\mu L} \]
- For the 6th dark fringe in vacuum:
\[ y_{6} = \frac{(2 \times 6 - 1) \lambda D}{2} = \frac{11 \lambda D}{2} \]

4. **Equating the Two Positions**:
- Since both fringes lie at the same position on the screen, we can set \( y_{10} = y_{6} \):
\[ \frac{10 \lambda D}{\mu L} = \frac{11 \lambda D}{2} \]

5. **Cancelling Common Terms**:
- We can cancel \( \lambda D \) from both sides (assuming \( \lambda D \neq 0 \)):
\[ \frac{10}{\mu L} = \frac{11}{2} \]

6. **Solving for µL**:
- Rearranging the equation gives:
\[ \mu L = \frac{10 \times 2}{11} = \frac{20}{11} \]

7. **Calculating the Approximate Value**:
- Now, calculating \( \frac{20}{11} \):
\[ \mu L \approx 1.818 \approx 1.8 \]

### Final Answer:
The refractive index of the liquid (µL) is approximately **1.8**.

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