A monochromatic light beam of frequency `6.0xx10^14 Hz` crosses from air into a transparent material where its wavelength is measured to be `300 nm.` What is the index of refraction of the material?

Explanation

To find the index of refraction of the material, we can use the relationship between the speed of light in vacuum (or air), the speed of light in the material, and the index of refraction. The formula we will use is:

\[ n = \frac{c}{v} \]

Where:
- \( n \) is the index of refraction,
- \( c \) is the speed of light in vacuum (approximately \( 3.0 \times 10^8 \) m/s),
- \( v \) is the speed of light in the material.

We can also relate the speed of light in the material to its frequency and wavelength using the equation:

\[ v = f \cdot \lambda \]

Where:
- \( f \) is the frequency of the light,
- \( \lambda \) is the wavelength of the light in the material.

Given:
- Frequency \( f = 6.0 \times 10^{14} \) Hz,
- Wavelength \( \lambda = 300 \) nm = \( 300 \times 10^{-9} \) m.

### Step 1: Calculate the speed of light in the material \( v \)

Using the formula \( v = f \cdot \lambda \):

\[ v = (6.0 \times 10^{14} \, \text{Hz}) \cdot (300 \times 10^{-9} \, \text{m}) \]

Calculating this gives:

\[ v = 6.0 \times 300 \times 10^{14} \times 10^{-9} = 1800 \times 10^{5} \, \text{m/s} = 1.8 \times 10^{6} \, \text{m/s} \]

### Step 2: Calculate the index of refraction \( n \)

Now that we have \( v \), we can find the index of refraction using:

\[ n = \frac{c}{v} \]

Substituting the values:

\[ n = \frac{3.0 \times 10^8 \, \text{m/s}}{1.8 \times 10^6 \, \text{m/s}} \]

Calculating this gives:

\[ n = \frac{3.0}{1.8} \times 10^{8-6} = \frac{3.0}{1.8} \times 10^{2} = 1.67 \]

### Final Answer

The index of refraction of the material is:

\[ n \approx 1.67 \]

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