Calculate Infrared Wave Wavelength From Frequency

by Alex Johnson 50 views

Hey there, physics enthusiasts! Ever wondered about the relationship between the frequency of a wave and its wavelength, especially when it's zipping through the vastness of a vacuum? Today, we're diving deep into a classic physics problem: calculating the wavelength of an infrared wave given its frequency. This isn't just about crunching numbers; it's about understanding the fundamental properties of electromagnetic radiation. We'll break down the concepts, walk through the calculation step-by-step, and explore why this matters in the world of physics. So, grab your thinking caps, and let's unravel the mysteries of light waves!

Understanding the Wave Equation: The Core Concept

The fundamental relationship that governs wave motion, especially electromagnetic waves like infrared radiation, is the wave equation. This equation beautifully connects three key properties of a wave: its speed, its frequency, and its wavelength. For any wave traveling through a vacuum, its speed is constant – it's the speed of light, denoted by the symbol 'cc'. This universal speed limit is approximately 3.00imes1083.00 imes 10^8 meters per second. Frequency, measured in Hertz (Hz), tells us how many wave cycles pass a point in one second. Think of it as the 'shakiness' or oscillation rate of the wave. Wavelength, often represented by the Greek letter lambda ('$ \lambda

), is the spatial period of the wave – essentially, the distance between two consecutive crests or troughs. Understanding this equation is key to solving problems involving waves, whether they are sound waves, water waves, or light waves traveling through space.

The Speed of Light: A Universal Constant

When we talk about waves traveling through a vacuum, the most crucial constant we rely on is the speed of light, 'cc'. This isn't just any speed; it's the ultimate speed limit in the universe according to our current understanding of physics. Its value is precisely defined as 299,792,458299,792,458 meters per second, but for most calculations, we use the approximation 'c≈3.00×108c \approx 3.00 \times 10^8 m/s'. This constant speed is intrinsic to the nature of electromagnetic waves, which include radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Because these waves don't need a medium to travel, they can propagate through the emptiness of space at this astonishing speed. Knowing 'cc' is essential because it acts as the bridge connecting frequency and wavelength in our calculations. Without this constant, we wouldn't be able to quantify how 'spread out' or 'compressed' a wave is based on how often it oscillates.

Frequency vs. Wavelength: An Inverse Relationship

Here's where the magic happens: the speed of light, frequency, and wavelength are intrinsically linked. The equation that ties them all together is:

c=f×λc = f \times \lambda

Where:

This equation reveals a critical inverse relationship between frequency and wavelength. If the speed of light 'cc' is constant, then as the frequency 'ff' increases, the wavelength '$ \lambda

must decrease, and vice versa. Imagine a string being shaken. If you shake it faster (higher frequency), the waves you create will be closer together (shorter wavelength). If you shake it slower (lower frequency), the waves will be more spread out (longer wavelength). This principle is fundamental to understanding the electromagnetic spectrum. High-frequency waves, like gamma rays, have very short wavelengths, while low-frequency waves, like radio waves, have very long wavelengths. Infrared radiation sits in a specific part of this spectrum, characterized by its particular range of frequencies and corresponding wavelengths.

Applying the Formula to Our Infrared Wave Problem

Now, let's get practical and apply this knowledge to the specific problem at hand. We are given an infrared wave traveling through a vacuum with a frequency of 4.0imes10144.0 imes 10^{14} Hz. Our goal is to determine its wavelength. To do this, we need to rearrange the fundamental wave equation to solve for wavelength. Remember the equation: 'c=f×λc = f \times \lambda'. To isolate '$ \lambda ′,wesimplydividebothsidesby′', we simply divide both sides by 'f

:

$ \lambda = c / f $

This rearranged formula is our tool for finding the wavelength. We know the speed of light 'cc' is approximately 3.00imes1083.00 imes 10^8 m/s, and we are given the frequency 'ff' as 4.0imes10144.0 imes 10^{14} Hz. Plugging these values into the formula will give us the answer we're looking for. It's important to ensure our units are consistent. Since 'cc' is in meters per second and 'ff' is in Hertz (which is equivalent to cycles per second or 1/s1/s), the resulting wavelength will be in meters, which is exactly what we need.

Step-by-Step Calculation

Let's break down the calculation:

  1. Identify the knowns:

  2. Identify the unknown:

  3. Use the rearranged formula:

  4. Substitute the values:

  5. Perform the division:

  6. Combine the results:

  7. Convert to standard scientific notation:

So, the wavelength of the infrared wave is 7.5imes10−77.5 imes 10^{-7} meters. This is a very small distance, which is characteristic of infrared light.

Analyzing the Options Provided

We've performed our calculation and arrived at a wavelength of 7.5imes10−77.5 imes 10^{-7} meters. Now, let's look at the multiple-choice options provided to see which one matches our result:

Comparing our calculated value, 7.5imes10−77.5 imes 10^{-7} m, directly with the options, we can see that Option B is the correct answer. Options A, C, and D represent significantly different magnitudes, likely resulting from calculation errors such as incorrect manipulation of exponents or using the wrong constant. For instance, Option A has a positive exponent, suggesting a vastly larger wavelength, while Options C and D involve different numerical values and exponent calculations that don't align with our formula. It's always a good practice to double-check your calculations, especially the exponent arithmetic, to avoid these common pitfalls.

Why is This Important? The Significance of Wavelength and Frequency

Understanding the relationship between frequency and wavelength, and being able to calculate them, is fundamental to numerous fields in physics and technology. Infrared radiation itself plays a crucial role in our daily lives and in scientific research. It's the type of electromagnetic radiation we feel as heat. When you stand near a fire, the warmth you feel is primarily due to infrared waves emitted by the flames. Beyond just heat, infrared imaging is used in night vision devices, thermal cameras for detecting heat leaks in buildings or for medical diagnostics, and in remote sensing for environmental monitoring. The specific wavelength of infrared radiation determines its properties and how it interacts with matter. For example, different wavelengths of infrared light are absorbed or reflected differently by various materials, which is the principle behind many spectroscopic techniques used to identify substances. The precise wavelength dictates its energy as well, with shorter wavelengths (higher frequencies) carrying more energy.

The Electromagnetic Spectrum: A Visual Guide

To truly appreciate where infrared radiation fits in, let's visualize the electromagnetic spectrum. It's a continuum of all electromagnetic waves arranged in order of increasing frequency and decreasing wavelength (or vice versa). Here's a simplified overview:

Type of Radiation Approximate Frequency Range (Hz) Approximate Wavelength Range (m)
Radio Waves <3imes1011< 3 imes 10^{11} >1> 1
Microwaves 3imes1011−3imes10143 imes 10^{11} - 3 imes 10^{14} 1imes10−3−11 imes 10^{-3} - 1
Infrared 3imes1014−4.3imes10143 imes 10^{14} - 4.3 imes 10^{14} 7imes10−7−1imes10−37 imes 10^{-7} - 1 imes 10^{-3}
Visible Light 4.3imes1014−7.5imes10144.3 imes 10^{14} - 7.5 imes 10^{14} 4imes10−7−7imes10−74 imes 10^{-7} - 7 imes 10^{-7}
Ultraviolet (UV) 7.5imes1014−3imes10167.5 imes 10^{14} - 3 imes 10^{16} 1imes10−8−4imes10−71 imes 10^{-8} - 4 imes 10^{-7}
X-rays 3imes1016−3imes10193 imes 10^{16} - 3 imes 10^{19} 1imes10−11−1imes10−81 imes 10^{-11} - 1 imes 10^{-8}
Gamma Rays >3imes1019> 3 imes 10^{19} <1imes10−11< 1 imes 10^{-11}

Our calculated infrared wave has a frequency of 4.0imes10144.0 imes 10^{14} Hz. Looking at the table, this frequency falls squarely within the infrared region. The corresponding wavelength we found, 7.5imes10−77.5 imes 10^{-7} m, is also right at the edge of the infrared spectrum, bordering visible red light (which typically starts around 7.5imes10−77.5 imes 10^{-7} m). This highlights how finely tuned the electromagnetic spectrum is, with adjacent regions having very similar properties.

Applications in Technology and Science

The ability to calculate and understand wavelengths is not just an academic exercise; it has profound practical implications. In telecommunications, the wavelength of radio waves and microwaves dictates the design of antennas and the bandwidth available for transmitting information. In medical imaging, the wavelength of X-rays is critical for their ability to penetrate tissues and create diagnostic images. For infrared, understanding its specific wavelengths is key to developing advanced thermal sensors, spectroscopy for chemical analysis, and even in astronomy for observing cooler objects in space that emit primarily in the infrared. Each part of the electromagnetic spectrum, defined by its frequency and wavelength, has unique applications and properties that scientists and engineers leverage to solve problems and innovate.

Frequently Asked Questions (FAQ)

Here are some common questions you might have about calculating wave properties:

Conclusion

In conclusion, calculating the wavelength of an electromagnetic wave from its frequency is a straightforward application of the fundamental wave equation, c=f×λc = f \times \lambda. For our infrared wave traveling through a vacuum with a frequency of 4.0imes10144.0 imes 10^{14} Hz, we used the constant speed of light (c≈3.00imes108c \approx 3.00 imes 10^8 m/s) and the rearranged formula $ \lambda = c / f $ to find its wavelength. The calculation yielded a wavelength of 7.5imes10−77.5 imes 10^{-7} meters, making option B the correct answer. This exercise not only reinforces our understanding of wave properties but also highlights the vast and fascinating nature of the electromagnetic spectrum and its critical role in science and technology. Keep exploring, keep questioning, and keep calculating!