Problem 2134 - JEE Physics

Question

The frequency of electromagnetic wave having wavelength \(25 \mathrm{~mm}\) is \(\quad \mathrm{Hz}\) (A) \(1.2 \times \overline{10^{10}}\) (B) \(7.5 \times 10^{5}\) (C) \(1.2 \times 10^{8}\) (D) \(7.5 \times 10^{6}\)

Step-by-Step Solution

Answer

The frequency of an electromagnetic wave having wavelength \(25~mm\) is \(1.2 \times 10^{10}~Hz\). The correct answer is (A).

Step 1: Write the given values

We are given the wavelength \(\lambda = 25~mm\) or \(2.5 \times 10^{-2}~m\) (remember to convert millimeters to meters). The speed of light \(c = 3 \times 10^8 m/s\).

Step 2: Use the formula to find the frequency

We will now use the formula \(f = \frac{c}{\lambda}\) to find the frequency: \(f = \frac{3 \times 10^8~m/s}{2.5 \times 10^{-2}~m}\)

Step 3: Perform the calculation

Now we will proceed to perform the calculation: \(f = \frac{3 \times 10^8}{2.5 \times 10^{-2}}\) \(f = \frac{3}{2.5} \times \frac{10^8}{10^{-2}}\) \(f = 1.2 \times 10^{10}~Hz\)

Step 4: Identify the correct option

Comparing our obtained frequency, \(f = 1.2 \times 10^{10}~Hz\), to the given options, the correct answer is: (A) \(1.2 \times 10^{10}~Hz\)

Key Concepts

Wavelength

Understanding the concept of wavelength is essential when studying electromagnetic waves. Wavelength, often symbolized as \( \lambda \), is the distance between successive crests (or troughs) of a wave. It is typically measured in meters (although other units like millimeters or nanometers may be used based on the context). Wavelength is a core characteristic that helps us identify different types of electromagnetic waves, such as radio waves, visible light, or gamma rays. In the exercise provided, the wavelength is given as \( 25 \text{ mm} \). It is important to convert it into meters for calculational purposes, so \( 25 \text{ mm} \) becomes \( 2.5 \times 10^{-2} \text{ m} \).

• It is vital to convert units properly while dealing with wavelengths.
• In general, as wavelength increases, the frequency decreases for a constant speed like that of light.

Frequency Calculation

Frequency, denoted by \( f \), measures how many waves pass a point in one second. It is expressed in Hertz (Hz). Calculating frequency involves dividing the speed of the wave by the wavelength:\[ f = \frac{c}{\lambda} \]Where \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)), and \( \lambda \) is the wavelength in meters. In the provided example:

• Given: \( c = 3 \times 10^8 \text{ m/s} \), \( \lambda = 2.5 \times 10^{-2} \text{ m} \)
• Frequency calculation: \( f = \frac{3 \times 10^8}{2.5 \times 10^{-2}} = 1.2 \times 10^{10} \text{ Hz} \)

Understanding frequency is crucial because it helps differentiate electromagnetic waves, much like how wavelength does. A higher frequency means more waves passing through a point per second, which is often associated with higher energy in electromagnetic waves.

Speed of Light

The speed of light, represented by \( c \), is one of the fundamental constants in physics—\( 3 \times 10^8 \text{ m/s} \). This constant describes how fast electromagnetic waves travel through a vacuum.Knowing the speed of light is crucial for a variety of calculations involving electromagnetic waves, particularly when converting between wavelength and frequency. For almost all questions regarding electromagnetic waves, any calculations assume that waves propagate in a vacuum unless specified otherwise. Thus, \( c = 3 \times 10^8 \text{ m/s} \) is used as a constant in equations.

• Electromagnetic waves, including visible light, travel at this constant speed in a vacuum.
• The speed of light is pivotal for understanding the relationship \( c = \lambda \times f \), where it connects wavelength and frequency.

This understanding helps solve problems relating to electromagnetic waves and is foundational for fields like optics and quantum physics.