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. `1.54`
C. `1.67`
D. `1.2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We know that in Young's double slit experiment, the position of bright and dark fringes can be determined using the formula for the distance from the center to the fringe on the screen. We need to find the refractive index of the liquid based on the given fringe conditions.

### Step 2: Write the formulas for bright and dark fringes
For the bright fringe in a medium (liquid):
\[ y_b = \frac{n \lambda D}{\mu d} \]
For the dark fringe in a medium (vacuum):
\[ y_d = \frac{(2n - 1) \lambda D}{2 d} \]

Where:
- \( y_b \) = distance of the bright fringe from the center
- \( y_d \) = distance of the dark fringe from the center
- \( n \) = order of the fringe
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits
- \( \mu \) = refractive index of the medium (liquid in this case)

### Step 3: Set up the equation based on the problem statement
According to the problem, the 10th bright fringe in liquid corresponds to the 6th dark fringe in vacuum. Therefore, we can equate the two distances:
\[ y_{10} = y_{6} \]

Substituting the formulas:
\[ \frac{10 \lambda D}{\mu_L d} = \frac{(2 \cdot 6 - 1) \lambda D}{2 d} \]

### Step 4: Simplify the equation
We can cancel \( \lambda D \) and \( d \) from both sides:
\[ \frac{10}{\mu_L} = \frac{11}{2} \]

### Step 5: Solve for the refractive index \( \mu_L \)
Rearranging the equation gives:
\[ \mu_L = \frac{10 \cdot 2}{11} = \frac{20}{11} \]

### Step 6: Calculate the approximate value
Calculating the numerical value:
\[ \mu_L \approx 1.818 \]

Thus, the refractive index of the liquid is approximately:
\[ \mu_L \approx 1.8 \]

### Final Answer
The refractive index of the liquid is approximately **1.8**.

---