Problem 2136 - JEE Physics

Question

What is the ratio of velocities of light rays of wavelengths \(4000^{\circ} \mathrm{A}\) and \(8000^{\circ} \mathrm{A}\) in vacuum? (A) \(1: 2\) (B) \(1: 1\) (C) \(2: 1\) (D) cannot be determined

Step-by-Step Solution

Answer

The ratio between the velocities of light rays with given wavelengths is equal to the ratio of their frequencies. We find that the ratio of frequencies for the wavelengths \(4000^{\circ} \mathrm{A}\) and \(8000^{\circ} \mathrm{A}\) is \(2:1\). Therefore, the correct answer is (C) \(2:1\).

Step 1: Determine the relationship between speed, frequency, and wavelength of light in vacuum.

The relationship between the speed of light (\(v\)), its frequency (\(f\)), and its wavelength (\(\lambda\)) in vacuum is given by the formula: \(v = f\lambda\)

Step 2: Identify the constant speed of light in vacuum.

Regardless of the wavelength of light - the speed of light (\(c\)) in vacuum is always constant, with the value of: \(c = 3 \times 10^8 \ m/s\)

Step 3: Calculate the frequency for the given wavelengths.

We want to find the ratio of velocities of light rays with wavelengths \(\lambda_1 = 4000^{\circ} \mathrm{A}\) and \(\lambda_2 = 8000^{\circ} \mathrm{A}\). Using the relationship \(v = f\lambda\), we have: \(f_1 = \frac{c}{\lambda_1}\) and \(f_2 = \frac{c}{\lambda_2}\)

Step 4: Find the ratio of frequencies.

Now, let's find the ratio of frequencies \(f_1\) and \(f_2\). Here, we have: \(\frac{f_1}{f_2} = \frac{\frac{c}{\lambda_1}}{\frac{c}{\lambda_2}} = \frac{\lambda_2}{\lambda_1} = \frac{8000^{\circ} \mathrm{A}}{4000^{\circ} \mathrm{A}} = \frac{8000}{4000} = 2\)

Step 5: Confirm the answer using the given options.

The ratio between the velocities of light rays with given wavelengths is equal to the ratio of their frequencies, which we found as \(2:1\). Therefore, the correct answer is (C) \(2:1\).

Key Concepts

Wavelength Dependency of Frequency

Light behaves in fascinating ways, and understanding the relationship between its frequency and wavelength is key to many areas of science. When discussing "wavelength dependency of frequency," we are exploring how changing the wavelength affects the frequency of light.

In scientific terms:

• The frequency (\(f\)) of light is inversely proportional to its wavelength (\(\lambda\)).
• This means that if the wavelength is doubled, the frequency is halved.

This relationship is crucial because, in a vacuum, the speed of light remains constant, so any change in wavelength has a direct impact on frequency.

For example, when light has wavelengths of \(4000^{\circ} \mathrm{A}\) and \(8000^{\circ} \mathrm{A}\), the frequency of the latter will be half of the former. By understanding this connection, we can predict how light behaves in different scenarios.

Constant Speed of Light

The speed of light in a vacuum is one of the most important constants in physics. It is denoted by the symbol \(c\) and has a fixed value of \(3 \times 10^8 \, \text{m/s}\).

This constancy is a fundamental principle because:

• Light travels at the same speed regardless of its wavelength or frequency in a vacuum.
• This constant speed is the baseline for many scientific calculations and theories.

The unchanging nature of light's speed ensures that equations like \(v = f\lambda\) are reliable and can be universally applied. It means that even if you alter the wavelength, the speed remains unchanged, shaping our understanding of light and electromagnetic waves.

Frequency-Wavelength Relationship

The relationship between frequency and wavelength is mathematically expressed by the equation \(v = f\lambda\). Here, \(v\) represents the speed of light, \(f\) is the frequency, and \(\lambda\) is the wavelength.

In a vacuum:

• The equation simplifies to \(c = f\lambda\), because the speed of light is constant.
• This shows a direct relationship between frequency and wavelength: as one increases, the other decreases proportionally.

By understanding this link, you can easily determine the properties of light by rearranging the equation. For instance, to find the frequency, use \(f = \frac{c}{\lambda}\). This versatility explains why different colors and types of light, with varying wavelengths, all travel at the same speed in a vacuum.