Problem 56 - Master Resource Book in JEE Main Physics

Question

A light of wavelength \(6000 \AA\) in air enters a medium of refractive index \(1.5 .\) Inside the medium, its frequency is \(v\) and its wavelength is \(\lambda\). Then, (a) \(\mathrm{v}=5 \times 10^{14} \mathrm{~Hz}\) (b) \(v=7.5 \times 10^{14} \mathrm{~Hz}\) (c) \(\lambda=4000 \AA\) (d) \(\lambda=9000 \dot{A}\)

Step-by-Step Solution

Answer

(a) \(v=5 \times 10^{14} \text{ Hz}\) and (c) \(\lambda=4000 \dot{A}\) are correct.

Step 1: Calculate frequency in air

The frequency of light does not change when it enters a different medium. The frequency, \(v_0\), can be calculated using the formula \(v_0 = \frac{c}{\lambda_0}\), where \(c\) is the speed of light in air (\(3 \times 10^8 \text{ m/s}\)) and \(\lambda_0\) is the wavelength in air (\(6000 \dot{A} = 6000 \times 10^{-10} \text{ m}\)). This gives \(v_0 = \frac{3 \times 10^8}{6000 \times 10^{-10}} = 5 \times 10^{14} \text{ Hz}\).

Step 2: Verify frequency inside the medium

Since the frequency \(v\) of light does not change when it passes from one medium to another, the frequency inside the medium is also \(v = 5 \times 10^{14} \text{ Hz}\). This matches option (a).

Step 3: Calculate wavelength in the medium

The wavelength in the medium, \(\lambda\), is found using \(\lambda = \frac{\lambda_0}{n}\), where \(n\) is the refractive index of the medium (1.5) and \(\lambda_0\) is the original wavelength in air (\(6000 \dot{A}\)). Calculate \(\lambda = \frac{6000 \dot{A}}{1.5} = 4000 \dot{A}\). This matches option (c).

Key Concepts

Frequency Calculation

The frequency of light is a key property that remains unchanged as it travels through different media. This fundamental characteristic stems from the way light interacts with the medium it's passing through. To calculate the frequency of light in air, we rely on the formula:\[ v_0 = \frac{c}{\lambda_0} \]Where:

• \(v_0\) is the frequency of light in air
• \(c\) is the speed of light in air, approximately \(3 \times 10^8\text{ m/s}\)
• \(\lambda_0\) is the wavelength of the light in air

For a wavelength of 6000 Å (which is 6000 × 10^-10 meters) in air, the frequency calculates to:\[ v_0 = \frac{3 \times 10^8}{6000 \times 10^{-10}} = 5 \times 10^{14}\, \text{Hz} \] This frequency remains unchanged when light enters another medium, such as glass or water. Thus, inside the medium described in the exercise with a refractive index of 1.5, this frequency is still \(5 \times 10^{14}\, \text{Hz}\). This property ensures that many optical properties are preserved, aiding in applications like fiber optics.

Wavelength in Different Media

As light transitions between different materials, its speed and, therefore, its wavelength can change. This happens because light travels at different speeds in various media. Notably, the frequency remains constant during this transition. The relationship between the wavelength in a new medium and the original medium is described by:\[ \lambda = \frac{\lambda_0}{n} \]Where:

• \(\lambda\) is the wavelength of light in the new medium
• \(\lambda_0\) is the wavelength of light in the original medium
• \(n\) is the refractive index of the new medium

In this exercise, the original wavelength \(\lambda_0\) is 6000 Å. The medium has a refractive index of 1.5. Placing these values into the formula gives:\[ \lambda = \frac{6000 \dot{A}}{1.5} = 4000 \dot{A} \]This calculation shows that the wavelength decreases as light enters a denser medium. Understanding this contraction is crucial for designing lenses and optical instruments, where precise control of light is required.

Refractive Index

The refractive index, often denoted as \(n\), is a fundamental property of materials that describes how much light is slowed down as it enters the material. Essentially, it measures the change in speed of light from a vacuum (or air) to another medium. The refractive index is typically greater than one for materials like glass or water:

• \(n = \frac{c}{v}\)
• \(c\) is the speed of light in a vacuum
• \(v\) is the speed of light in the medium

The refractive index indicates how much the path of light is bent, or refracted, when entering a material. In the exercise, the medium's refractive index is 1.5, signifying that light travels slower in this medium than in a vacuum.
When light passes from one medium to another, such as air to water, the change in speed results in a deviation in its path—often observable as a bent straw in a glass of water. This concept does not only help in understanding natural phenomena but also plays a vital role in engineering and technological advancements, such as designing eyeglasses to improve vision.