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Physics: Waves, Sound, and Light - Electromagnetic Spectrum

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Waves are oscillations that transfer energy without transferring matter. They are categorized into mechanical waves (require a medium) and electromagnetic (EM) waves (do not require a medium).

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Electromagnetic waves are transverse waves consisting of oscillating electric and magnetic fields. In a vacuum, all EM waves travel at the constant speed of light, c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}.

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The Electromagnetic Spectrum is the range of all types of EM radiation, ordered by frequency (ff) or wavelength (λ\lambda). The order from longest wavelength (lowest frequency) to shortest wavelength (highest frequency) is: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, and Gamma rays.

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Wavelength (λ\lambda) is the distance between two consecutive crests or troughs, measured in meters (mm).

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Frequency (ff) is the number of wave cycles passing a point per second, measured in Hertz (HzHz). The relationship is inverse: as frequency increases, wavelength decreases (f∝1λf \propto \frac{1}{\lambda}).

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The energy (EE) of an EM wave is directly proportional to its frequency. Therefore, Gamma rays carry the most energy and are the most ionizing/dangerous, while Radio waves carry the least energy.

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Visible light is the only part of the spectrum detectable by the human eye, ranging from Red (longest λ\lambda, lowest ff) to Violet (shortest λ\lambda, highest ff), often remembered by the acronym VIBGYOR.

📐Formulae

v=fλv = f \lambda

c=fλc = f \lambda

f=1Tf = \frac{1}{T}

c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

A radio station transmits waves with a wavelength of λ=150 m\lambda = 150 \text{ m}. Calculate the frequency of these radio waves, assuming they travel at the speed of light c=3×108 m/sc = 3 \times 10^8 \text{ m/s}.

Solution:

Using the wave equation c=fλc = f \lambda, we rearrange to solve for frequency: f=cλf = \frac{c}{\lambda}. Substituting the values: f=3×108 m/s150 mf = \frac{3 \times 10^8 \text{ m/s}}{150 \text{ m}} f=2×106 Hzf = 2 \times 10^6 \text{ Hz} or 2 MHz2 \text{ MHz}.

Explanation:

Since radio waves are part of the electromagnetic spectrum, they travel at the speed of light. Dividing the speed by the wavelength gives the number of cycles per second.

Problem 2:

An electromagnetic wave has a frequency of 6×1014 Hz6 \times 10^{14} \text{ Hz}. Determine its wavelength and identify which part of the visible spectrum it likely belongs to (Red ≈700 nm\approx 700 \text{ nm}, Violet ≈400 nm\approx 400 \text{ nm}), given 1 nm=10−9 m1 \text{ nm} = 10^{-9} \text{ m}.

Solution:

Using λ=cf\lambda = \frac{c}{f}: λ=3×1086×1014\lambda = \frac{3 \times 10^8}{6 \times 10^{14}} λ=0.5×10−6 m\lambda = 0.5 \times 10^{-6} \text{ m} λ=500×10−9 m=500 nm\lambda = 500 \times 10^{-9} \text{ m} = 500 \text{ nm}.

Explanation:

The wavelength is 500 nm500 \text{ nm}. Since this falls between 400 nm400 \text{ nm} and 700 nm700 \text{ nm}, it is visible light, specifically in the blue-green range.