A light of wavelength 6000 in air enters a medium of refractive index `1.5`. Inside the medium , its frequency is v and it wavelength is `lambda`.

Options

A. `v=5xx10^(14)Hz`
B. `v=7.5xx10^(14)Hz`
C. `lambda=4000`
D. `A=9000`

Answer

To solve the problem step by step, we will determine the frequency and wavelength of light as it enters a medium with a refractive index of 1.5.

### Step 1: Understand the relationship between speed, frequency, and wavelength
The speed of light in a medium is given by the equation:
\[ c = \nu \cdot \lambda \]
where:
- \( c \) is the speed of light,
- \( \nu \) is the frequency,
- \( \lambda \) is the wavelength.

### Step 2: Convert the given wavelength from angstroms to meters
The given wavelength in air is 6000 angstroms. We convert this to meters:
\[ 6000 \, \text{angstroms} = 6000 \times 10^{-10} \, \text{meters} = 6 \times 10^{-7} \, \text{meters} \]

### Step 3: Calculate the frequency in air
Using the speed of light in air, which is approximately \( 3 \times 10^8 \, \text{m/s} \), we can find the frequency:
\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{6 \times 10^{-7} \, \text{m}} \]
Calculating this gives:
\[ \nu = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} \, \text{Hz} = 5 \times 10^{14} \, \text{Hz} \]

### Step 4: Determine that frequency remains constant
The frequency of light does not change when it enters a different medium. Therefore, the frequency in the medium (with refractive index 1.5) is also:
\[ \nu = 5 \times 10^{14} \, \text{Hz} \]

### Step 5: Calculate the speed of light in the medium
The speed of light in a medium can be calculated using the refractive index \( n \):
\[ v = \frac{c}{n} \]
where \( n = 1.5 \). Thus:
\[ v = \frac{3 \times 10^8 \, \text{m/s}}{1.5} = 2 \times 10^8 \, \text{m/s} \]

### Step 6: Calculate the wavelength in the medium
Now we can find the wavelength in the medium using the frequency we calculated earlier:
\[ \lambda = \frac{v}{\nu} = \frac{2 \times 10^8 \, \text{m/s}}{5 \times 10^{14} \, \text{Hz}} \]
Calculating this gives:
\[ \lambda = 0.4 \times 10^{-6} \, \text{m} = 4 \times 10^{-7} \, \text{m} = 4000 \, \text{angstroms} \]

### Summary of Results
- Frequency in the medium: \( 5 \times 10^{14} \, \text{Hz} \)
- Wavelength in the medium: \( 4000 \, \text{angstroms} \)