A monochromatic light wave with wavelength `lambda_1` and frequency `ν_1` in air enters another medium. If the angle of incidence and angle of refraction at the interface are `45^@` and `30^@` respectively, then the wavelength `lambda_2` and frequency `ν_2` of the refracted wave are:

Options

A. `lambda_2 =sqrt(2lambda_1),v_2 =v_1`
B. `lambda_2 = lambda_1, v_2 = 1/(sqrt2)v_1`
C. `lambda_2 =lambda_1, v_2 = sqrt2 v_1`
D. `lambda_2 = 1/sqrt2 lambda_1,v_2 = v_1`

Answer

To solve the problem, we need to analyze the behavior of a monochromatic light wave as it transitions from air into another medium. We will use Snell's Law and the relationships between wavelength, frequency, and speed of light.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength in air: \( \lambda_1 \)
- Frequency in air: \( \nu_1 \)
- Angle of incidence: \( \theta_1 = 45^\circ \)
- Angle of refraction: \( \theta_2 = 30^\circ \)

2. **Use Snell's Law:**
Snell's Law states that:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
In air, the refractive index \( n_1 \) is approximately 1. Thus, we can write:
\[ \sin(45^\circ) = n_2 \sin(30^\circ) \]

3. **Calculate Sine Values:**
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

4. **Substitute Values into Snell's Law:**
\[ 1 \cdot \frac{1}{\sqrt{2}} = n_2 \cdot \frac{1}{2} \]
Rearranging gives:
\[ n_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

5. **Relate Frequencies:**
The frequency of light does not change when it enters a different medium. Therefore:
\[ \nu_2 = \nu_1 \]

6. **Relate Wavelengths:**
The relationship between the speed of light, frequency, and wavelength is given by:
\[ v = \nu \lambda \]
In the new medium, the speed of light \( v_2 \) is given by:
\[ v_2 = \frac{c}{n_2} \]
where \( c \) is the speed of light in vacuum. Thus:
\[ v_2 = \frac{c}{\sqrt{2}} \]
Since \( \nu_2 = \nu_1 \), we can relate the wavelengths:
\[ \lambda_2 = \frac{v_2}{\nu_2} = \frac{c/\sqrt{2}}{\nu_1} \]
And since \( v_1 = c \) (in air):
\[ \lambda_1 = \frac{c}{\nu_1} \]
Therefore:
\[ \lambda_2 = \frac{\lambda_1}{\sqrt{2}} \]

### Final Results:
- Frequency in the new medium: \( \nu_2 = \nu_1 \)
- Wavelength in the new medium: \( \lambda_2 = \frac{\lambda_1}{\sqrt{2}} \)