Light of wavelength `lambda_(0)` in air enters a medium of refractive index n. If two points A and B in this medium lie along the path of this light at a distance x, then phase difference `phi_(0)` between these two point is

Options

A. `phi_(0) =1/n(2pi)/(lambda_(0))x`
B. `phi_(0) =n(2pi)/lambda_(0)x`
C. `phi_(0) =(n-1)((2pi)/lambda_(0))x`
D. `phi_(0) = 1/(n-1)(2pi)/(lambda_(0))x`

Answer

To find the phase difference \( \phi_0 \) between two points A and B in a medium with refractive index \( n \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Optical Path Difference (OPD)**:
The optical path difference between two points A and B in a medium is given by the formula:
\[ \text{OPD} = n \cdot x \]
where \( n \) is the refractive index of the medium and \( x \) is the distance between points A and B.

2. **Relate Optical Path Difference to Phase Difference**:
The phase difference \( \phi_0 \) can be related to the optical path difference using the formula:
\[ \phi_0 = \frac{2\pi}{\lambda} \cdot \text{OPD} \]
where \( \lambda \) is the wavelength of light in the medium.

3. **Substitute the OPD into the Phase Difference Formula**:
By substituting the expression for OPD into the phase difference formula, we get:
\[ \phi_0 = \frac{2\pi}{\lambda} \cdot (n \cdot x) \]

4. **Adjust for Wavelength in Air**:
Since the wavelength of light in the medium is different from that in air, we can express the wavelength in the medium \( \lambda \) as:
\[ \lambda = \frac{\lambda_0}{n} \]
where \( \lambda_0 \) is the wavelength of light in air.

5. **Final Expression for Phase Difference**:
Substituting \( \lambda \) into the phase difference formula gives:
\[ \phi_0 = \frac{2\pi}{\frac{\lambda_0}{n}} \cdot (n \cdot x) = \frac{2\pi n x}{\lambda_0} \]

Thus, the phase difference \( \phi_0 \) between points A and B is:
\[ \phi_0 = \frac{2\pi n x}{\lambda_0} \]