A monochromatic beam of light of wavelength `6000 A` in vacuum enters a medium of refractive index `1.5`. In the medium its wavelength is…., its frequency is…..

Explanation

To solve the problem, we need to find the wavelength and frequency of a monochromatic beam of light when it enters a medium with a refractive index of 1.5. The initial wavelength of the light in vacuum is given as 6000 Å (angstroms).

### Step 1: Understand the relationship between wavelength, frequency, and speed of light
The speed of light in a medium (V) is related to the frequency (f) and wavelength (λ) by the equation:
\[ V = f \cdot \lambda \]

### Step 2: Determine the speed of light in the medium
The speed of light in a medium can be calculated using the refractive index (n):
\[ n = \frac{c}{V} \]
Where:
- \( c \) is the speed of light in vacuum (approximately \( 3 \times 10^8 \) m/s)
- \( V \) is the speed of light in the medium

From this, we can express the speed of light in the medium as:
\[ V = \frac{c}{n} \]

### Step 3: Calculate the speed of light in the medium
Given that the refractive index \( n = 1.5 \):
\[ V = \frac{3 \times 10^8 \text{ m/s}}{1.5} = 2 \times 10^8 \text{ m/s} \]

### Step 4: Calculate the frequency of the light
The frequency of light does not change when it enters a different medium. It can be calculated using the initial wavelength in vacuum:
\[ f = \frac{c}{\lambda_1} \]
Where:
- \( \lambda_1 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)

Now, substituting the values:
\[ f = \frac{3 \times 10^8 \text{ m/s}}{6000 \times 10^{-10} \text{ m}} \]
\[ f = \frac{3 \times 10^8}{6 \times 10^{-7}} \]
\[ f = 5 \times 10^{14} \text{ Hz} \]

### Step 5: Calculate the wavelength in the medium
Using the relationship between speed, frequency, and wavelength in the medium:
\[ V = f \cdot \lambda_2 \]
Where \( \lambda_2 \) is the wavelength in the medium. Rearranging gives:
\[ \lambda_2 = \frac{V}{f} \]

Substituting the values we have:
\[ \lambda_2 = \frac{2 \times 10^8 \text{ m/s}}{5 \times 10^{14} \text{ Hz}} \]
\[ \lambda_2 = 4 \times 10^{-7} \text{ m} = 4000 \, \text{Å} \]

### Final Results
- The wavelength in the medium \( \lambda_2 \) is **4000 Å**.
- The frequency \( f \) is **\( 5 \times 10^{14} \text{ Hz} \)**.