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 regarding Young's double slit experiment in a liquid, we need to analyze the relationship between the bright and dark fringes in both the liquid and vacuum. Here's a step-by-step solution:

### Step 1: Understand the fringe conditions
In Young's double slit experiment, the position of bright and dark fringes is given by the formulas:
- Bright fringe: \( y_n = n \frac{\lambda D}{d} \)
- Dark fringe: \( y_m = \left(m + \frac{1}{2}\right) \frac{\lambda D}{d} \)

Where:
- \( n \) is the order of the bright fringe
- \( m \) is the order of the dark fringe
- \( \lambda \) is the wavelength of light in vacuum
- \( D \) is the distance from the slits to the screen
- \( d \) is the distance between the slits

### Step 2: Relate the fringes in liquid and vacuum
Given that the 10th bright fringe in the liquid coincides with the 6th dark fringe in vacuum, we can set up the equations for both conditions.

For the 10th bright fringe in liquid:
\[ y_{10} = 10 \frac{\lambda'}{d} D \]
Where \( \lambda' = \frac{\lambda}{\mu} \) (the wavelength in the liquid).

For the 6th dark fringe in vacuum:
\[ y_{6} = \left(6 + \frac{1}{2}\right) \frac{\lambda}{d} D = \frac{13}{2} \frac{\lambda}{d} D \]

### Step 3: Equate the two positions
Since both positions correspond to the same point on the screen:
\[ 10 \frac{\lambda'}{d} D = \frac{13}{2} \frac{\lambda}{d} D \]

### Step 4: Simplify the equation
Cancel \( D \) and \( d \) from both sides:
\[ 10 \lambda' = \frac{13}{2} \lambda \]
Substituting \( \lambda' = \frac{\lambda}{\mu} \):
\[ 10 \left(\frac{\lambda}{\mu}\right) = \frac{13}{2} \lambda \]

### Step 5: Solve for the refractive index \( \mu \)
Now, we can solve for \( \mu \):
\[ 10 \frac{1}{\mu} = \frac{13}{2} \]
\[ \mu = \frac{10 \times 2}{13} = \frac{20}{13} \]

### Step 6: Approximate the refractive index
Calculating \( \frac{20}{13} \):
\[ \mu \approx 1.54 \]

### Final Answer
The refractive index of the liquid is approximately \( \mu \approx 1.54 \).

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