CIE AS/A Level Physics 7.4 Electromagnetic spectrum Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 7.4 Electromagnetic spectrum Study Notes – New Syllabus
CIE AS/A Level Physics 7.4 Electromagnetic spectrum Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- state that all electromagnetic waves are transverse waves that travel with the same speed c in free space
- recall the approximate range of wavelengths in free space of the principal regions of the electromagnetic spectrum from radio waves to γ-rays
- recall that wavelengths in the range 400–700 nm in free space are visible to the human eye
Nature and Speed of Electromagnetic Waves
Electromagnetic (EM) waves are transverse waves consisting of oscillating electric (\( \mathrm{\vec{E}} \)) and magnetic (\( \mathrm{\vec{B}} \)) fields that are perpendicular to each other and to the direction of wave propagation.
Key Characteristics:
- They do not require a medium — EM waves can travel through a vacuum.
- Both the electric and magnetic fields oscillate at right angles to the direction of travel.
- All electromagnetic waves propagate at the same speed in free space.
\( \mathrm{c = 3.0 \times 10^8 \ m/s} \)
Key Idea:
All electromagnetic waves — from radio waves to γ-rays — travel at the same speed \( \mathrm{c} \) in free space, regardless of their frequency or wavelength.
Transverse Nature of EM Waves:
- Electric field (\( \mathrm{\vec{E}} \)) oscillates in one plane.
- Magnetic field (\( \mathrm{\vec{B}} \)) oscillates in a plane perpendicular to \( \mathrm{\vec{E}} \).
- Wave travels in a direction perpendicular to both \( \mathrm{\vec{E}} \) and \( \mathrm{\vec{B}} \).
Relationship Between Speed, Frequency, and Wavelength:
\( \mathrm{c = f \lambda} \)
- \( \mathrm{c} \): speed of electromagnetic waves in free space (\( \mathrm{3.0 \times 10^8 \ m/s} \))
- \( \mathrm{f} \): frequency of the wave (Hz)
- \( \mathrm{\lambda} \): wavelength of the wave (m)
Key Takeaways:
- All EM waves are transverse in nature.
- They travel at the same speed \( \mathrm{c = 3.0 \times 10^8 \ m/s} \) in vacuum.
- They differ only in frequency and wavelength.
Example
The frequency of a radio wave is \( \mathrm{f = 100 \ MHz} \). Calculate its wavelength in free space.
▶️ Answer / Explanation
Given: \( \mathrm{f = 100 \times 10^6 \ Hz} \), \( \mathrm{c = 3.0 \times 10^8 \ m/s} \)
Using \( \mathrm{c = f \lambda} \):
\( \mathrm{\lambda = \dfrac{c}{f} = \dfrac{3.0 \times 10^8}{100 \times 10^6} = 3.0 \ m} \)
Therefore, the wavelength of the radio wave is 3.0 m.
Electromagnetic Spectrum and Wavelength Ranges
The electromagnetic spectrum is the complete range of all electromagnetic waves, arranged according to their wavelengths or frequencies.
Although all electromagnetic waves travel at the same speed \( \mathrm{c = 3.0 \times 10^8 \ m/s} \) in free space, they differ in their wavelength (\( \mathrm{\lambda} \)) and frequency (\( \mathrm{f} \)).
\( \mathrm{c = f \lambda} \)
Principal Regions of the Electromagnetic Spectrum:
| Region | Approximate Wavelength Range in Free Space | Common Source / Application |
|---|---|---|
| Radio waves | \( \mathrm{> 10^{-1} \ m} \) | Broadcasting, communication |
| Microwaves | \( \mathrm{10^{-3} \ m \ to \ 10^{-1} \ m} \) | Cooking, radar, satellite communication |
| Infrared (IR) | \( \mathrm{7 \times 10^{-7} \ m \ to \ 10^{-3} \ m} \) | Thermal imaging, remote controls |
| Visible light | \( \mathrm{4 \times 10^{-7} \ m \ to \ 7 \times 10^{-7} \ m} \) | Human vision, photography |
| Ultraviolet (UV) | \( \mathrm{10^{-8} \ m \ to \ 4 \times 10^{-7} \ m} \) | Fluorescent lamps, sterilization |
| X-rays | \( \mathrm{10^{-11} \ m \ to \ 10^{-8} \ m} \) | Medical imaging, security scanning |
| Gamma (γ) rays | \( \mathrm{< 10^{-11} \ m} \) | Nuclear reactions, cancer treatment |
Key Points:
- As wavelength decreases across the spectrum, frequency increases.
- Energy of EM waves increases with frequency (\( \mathrm{E = hf} \)).
- The electromagnetic spectrum extends over many orders of magnitude in wavelength and frequency.
Order of the Spectrum (Longest → Shortest Wavelength):
Example
Arrange the following electromagnetic waves in order of increasing frequency: ultraviolet, radio waves, X-rays, infrared.
▶️ Answer / Explanation
Frequency increases as wavelength decreases. Hence, the order of increasing frequency is:
Radio waves → Infrared → Ultraviolet → X-rays
Therefore, X-rays have the highest frequency, and radio waves have the lowest.
Visible Light and the Human Eye
The portion of the electromagnetic spectrum that can be detected by the human eye is known as visible light. It lies between the infrared and ultraviolet regions of the spectrum.
Wavelength Range of Visible Light:
\( \mathrm{400 \ nm \leq \lambda \leq 700 \ nm} \)
In free space, wavelengths within this range correspond to electromagnetic radiation that the human eye perceives as light of different colors.
Color and Wavelength Correspondence:
| Color | Approximate Wavelength (nm) | Approximate Frequency (×1014 Hz) |
|---|---|---|
| Violet | 400–450 | 7.5–7.0 |
| Blue | 450–495 | 6.7–6.0 |
| Green | 495–570 | 6.0–5.3 |
| Yellow | 570–590 | 5.3–5.1 |
| Orange | 590–620 | 5.1–4.8 |
| Red | 620–700 | 4.8–4.3 |
Key Features:
- Visible light is a very small part of the electromagnetic spectrum.
- Shorter wavelengths (≈400 nm) correspond to violet light, while longer wavelengths (≈700 nm) correspond to red light.
- Different colors correspond to different wavelengths, but all travel at the same speed \( \mathrm{c = 3.0 \times 10^8 \ m/s} \) in free space.
Applications:
- Human vision (sensitivity peaks around green light, ≈550 nm).
- Optical devices such as cameras, microscopes, and telescopes.
- Fiber optics communication using visible and near-infrared light.
Example
Determine the frequency of a green light wave with wavelength \( \mathrm{550 \ nm} \) in free space.
▶️ Answer / Explanation
Given: \( \mathrm{\lambda = 550 \ nm = 550 \times 10^{-9} \ m} \)
Using \( \mathrm{c = f \lambda} \), we have:
\( \mathrm{f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{550 \times 10^{-9}} = 5.45 \times 10^{14} \ Hz} \)
Therefore, the frequency of green light is approximately \( \mathrm{5.45 \times 10^{14} \ Hz} \).