Problem 55 - 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

Frequency in medium: \(5 \times 10^{14} \text{Hz}\). Wavelength in medium: \(4000 \text{\AA}\).

Step 1: Understanding the Relationship Between Frequency and Medium

The frequency of light does not change when it enters a different medium. Therefore, the frequency in the medium remains the same as in air. We will first calculate the frequency using the information given for the wavelength in air.

Step 2: Calculate the Frequency of Light

The speed of light in air is approximately \( c = 3 \times 10^8 \, \text{m/s} \). The wavelength in air is \( \lambda_0 = 6000 \, \text{\AA} = 6000 \times 10^{-10} \, \text{m} \). We use the formula for frequency: \[ u = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6000 \times 10^{-10}} \].By performing the calculation, we get \( u = 5 \times 10^{14} \, \text{Hz} \).

Step 3: Determine the Wavelength in the Medium

When light enters a medium with refractive index \( n \), its speed reduces to \( \frac{c}{n} \). The wavelength within the medium can be calculated using:\[ \lambda_{\text{medium}} = \frac{\lambda_0}{n} = \frac{6000}{1.5} \] Performing the division gives \( \lambda_{\text{medium}} = 4000 \, \text{\AA} \).

Key Concepts

Understanding Refractive Index

The refractive index is a crucial concept in optics. It measures how much a medium can bend or "refract" light. When light moves from one medium to another, its speed and wavelength change, but its frequency remains constant. The refractive index, represented by the symbol \( n \), is defined as the ratio of the speed of light in a vacuum \( c \) to the speed of light in the medium \( v \). This can be expressed as:

• \( n = \frac{c}{v} \)
• Where \( c = 3 \times 10^8 \text{ m/s} \)

A higher refractive index means the light slows down more in the medium, causing the light to bend more.
In our exercise, the given refractive index is 1.5, which implies that light travels 1.5 times faster in a vacuum than it does in this medium.

Wavelength Changes in a Medium

When light travels from one medium to another, its speed and wavelength are altered, but not its frequency. In any medium other than a vacuum, light slows down, shortening its wavelength. This is determined by the refractive index \( n \), using the following relationship:

• \( \lambda_{\text{medium}} = \frac{\lambda_0}{n} \)
• \( \lambda_0 \) is the initial wavelength in a vacuum or air

In our case, the initial wavelength \( \lambda_0 \) is \( 6000 \text{ \AA} \). Inside the medium with \( n = 1.5 \), the wavelength becomes \( 4000 \text{ \AA} \), calculated as \( \lambda_{\text{medium}} = \frac{6000}{1.5} \).
This reduction in wavelength happens because the speed of light has slowed within the medium. Despite this change in wavelength, the frequency of the light remains unchanged.

Frequency of Light in Different Media

Frequency is a fundamental property of light that doesn't change even when the light enters a different medium. The frequency of light is related to its speed and wavelength by the formula \( f = \frac{c}{\lambda_0} \). Since the speed of light \( c \) and the original wavelength \( \lambda_0 \) remain constant in vacuum or air, the frequency \( f \) calculated is also constant.

• In our example, the frequency \( f \) is \( 5 \times 10^{14} \text{ Hz} \).
• This is because the calculation \( f = \frac{3 \times 10^8}{6000 \times 10^{-10}} \) results in the same frequency regardless of the medium.

This property is essential in many applications of optics, as it means that frequency-based characteristics, like color in visible light, won't alter when entering different media. This constancy of frequency is vital in technologies like fiber optics, where light is transmitted through various materials.