A light wave in air enters a medium of refractive index `(4)/(3)`. If the wavelength of light in air is `6000Å`, then the wave number of light in the medium is

Options

A. `1.11 xx 10^(6) //m`
B. `2.22 xx 10^(6)//m`
C. `3.33 xx 10^(6)//m`
D. `4.44 xx 10^(6)//m`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the relationship between wavelength and refractive index
When light travels from one medium to another, its wavelength changes according to the refractive indices of the two media. The relationship is given by:
\[ \lambda_2 = \frac{\lambda_1}{\mu} \]
where:
- \(\lambda_1\) is the wavelength in the first medium (air),
- \(\lambda_2\) is the wavelength in the second medium,
- \(\mu\) is the refractive index of the second medium relative to the first.

### Step 2: Identify the given values
From the problem, we have:
- Wavelength in air, \(\lambda_1 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m}\)
- Refractive index of the medium, \(\mu = \frac{4}{3}\)

### Step 3: Calculate the wavelength in the medium
Using the relationship from Step 1:
\[ \lambda_2 = \frac{\lambda_1}{\mu} = \frac{6000 \times 10^{-10}}{\frac{4}{3}} = 6000 \times 10^{-10} \times \frac{3}{4} \]
Calculating this gives:
\[ \lambda_2 = 6000 \times 10^{-10} \times 0.75 = 4500 \times 10^{-10} \, \text{m} = 4500 \, \text{Å} \]

### Step 4: Determine the wave number in the medium
The wave number \(k\) is defined as:
\[ k = \frac{1}{\lambda} \]
For the medium, we have:
\[ k_2 = \frac{1}{\lambda_2} = \frac{1}{4500 \times 10^{-10}} = \frac{1}{4500 \times 10^{-10}} \, \text{m}^{-1} \]
Calculating this gives:
\[ k_2 = \frac{10^{10}}{4500} \approx 2.22 \times 10^6 \, \text{m}^{-1} \]

### Final Answer
The wave number of light in the medium is approximately:
\[ k_2 \approx 2.22 \times 10^6 \, \text{m}^{-1} \]

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