CIE AS/A Level Physics 8.2 Diffraction Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 8.2 Diffraction Study Notes – New Syllabus
CIE AS/A Level Physics 8.2 Diffraction 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:
- explain the meaning of the term diffraction
- show an understanding of experiments that demonstrate diffraction including the qualitative effect of the gap width relative to the wavelength of the wave; for example diffraction of water waves in a ripple tank
Diffraction of Waves
Diffraction is the phenomenon in which waves spread out as they pass through a gap (aperture) or around the edges of an obstacle. It is a fundamental property of all types of waves, including sound, light, and water waves.
Key Features:
- Diffraction occurs because each point on a wavefront acts as a secondary source of wavelets (Huygens’ Principle).
- When these secondary wavelets overlap, they form a new wavefront that appears to “bend” into the region beyond the obstacle or gap.
- The extent of diffraction depends on the relationship between the wavelength (\( \mathrm{\lambda} \)) and the size of the gap (\( \mathrm{a} \)).
Conditions for Significant Diffraction:
| Relative Size of Gap (a) and Wavelength (λ) | Observation | Extent of Diffraction |
|---|---|---|
| \( \mathrm{a \gg \lambda} \) | Waves pass almost straight through with little spreading. | Very small diffraction |
| \( \mathrm{a \approx \lambda} \) | Waves spread out widely after passing through the gap. | Maximum diffraction |
| \( \mathrm{a \ll \lambda} \) | Waves are strongly spread; may not pass directly through. | Large diffraction but lower transmitted intensity |
Key Idea:
Diffraction is most noticeable when the wavelength of the wave is comparable to the size of the gap or obstacle. Smaller wavelengths (like light) diffract much less than longer wavelengths (like sound or water waves).
Example
Explain why sound waves from a loudspeaker can be heard clearly even when the listener is around a corner, but light from the same source cannot be seen.
▶️ Answer / Explanation
Sound waves have much longer wavelengths (typically \( \mathrm{0.1 \ to \ 10 \ m} \)) compared to light waves (\( \mathrm{\approx 10^{-7} \ m} \)). Because the wavelength of sound is comparable to the size of the doorway or wall opening, it undergoes noticeable diffraction and spreads out into the region beyond the corner.
Light, on the other hand, has a wavelength millions of times smaller than the size of the doorway, so \( \mathrm{a \gg \lambda} \), leading to negligible diffraction. Thus, sound can bend around the corner, but light travels almost straight and cannot be seen.
Key Idea: The degree of diffraction increases as \( \mathrm{\dfrac{\lambda}{a}} \) increases — long-wavelength waves diffract more strongly than short-wavelength waves.
Experiments Demonstrating Diffraction
Diffraction can be observed in various types of waves. The following experiments demonstrate how it occurs and how the degree of diffraction changes with gap width.
| Wave Type | Experimental Setup | Observation | Conclusion |
|---|---|---|---|
| Water waves (ripple tank) | Plane waves are produced using a straight barrier with an adjustable gap. The pattern is observed on the ripple tank screen (or using light above the tank). | When the gap width \( \mathrm{a} \) is similar to the wavelength \( \mathrm{\lambda} \), the waves spread out after passing through the gap. | Confirms that diffraction increases as \( \mathrm{a} \) approaches \( \mathrm{\lambda} \). |
| Sound waves | A loudspeaker emits sound waves toward an opening or doorway. The sound can be heard even around a corner. | Sound waves (long wavelength) diffract widely around obstacles. | Shows that longer wavelengths produce greater diffraction. |
| Light waves | Light passes through a narrow slit or diffraction grating. A spreading (fringe) pattern is observed on a screen. | Noticeable diffraction occurs only when the slit width is very small — comparable to the light wavelength (~\( \mathrm{10^{-7} \ m} \)). | Demonstrates that visible light diffracts, but only through very narrow slits. |
Qualitative Effect of Gap Width on Diffraction:
- As the gap becomes narrower (approaching \( \mathrm{\lambda} \)), the spreading of waves increases.
- As the gap becomes wider compared to \( \mathrm{\lambda} \), diffraction decreases — waves pass almost straight through.
- For very large gaps (\( \mathrm{a \gg \lambda} \)), each part of the wavefront travels nearly straight, producing little curvature.
Key Formula (Qualitative Relationship):
\( \mathrm{Degree \ of \ diffraction \ \propto \ \dfrac{\lambda}{a}} \)
Example
In a ripple tank experiment, the wavelength of water waves is \( \mathrm{2.0 \ cm} \). Predict what happens when the waves encounter a gap of width
(a) 10 cm,
(b) 3 cm, and
(c) 2 cm.
▶️ Answer / Explanation
(a) \( \mathrm{a = 10 \ cm \gg \lambda} \): Only slight spreading occurs; waves pass almost straight through.
(b) \( \mathrm{a = 3 \ cm \approx \lambda} \): Significant spreading; waves emerge in a broad curved front — maximum diffraction.
(c) \( \mathrm{a = 2 \ cm = \lambda} \): Waves spread almost uniformly in all directions beyond the gap — strongest diffraction observed.
Example
In a ripple tank, plane water waves of wavelength \( \mathrm{2.5 \ cm} \) are directed toward a barrier with an adjustable gap. Describe the diffraction pattern observed when the gap width is (a) \( \mathrm{10 \ cm} \), (b) \( \mathrm{3 \ cm} \), and (c) \( \mathrm{2.5 \ cm} \).
▶️ Answer / Explanation
(a) \( \mathrm{a = 10 \ cm \gg \lambda} \): Waves pass almost straight through the gap with little spreading. The transmitted wavefront remains nearly flat — diffraction is minimal.
(b) \( \mathrm{a = 3 \ cm \approx \lambda} \): Waves spread out noticeably after the gap, forming curved wavefronts. The diffraction is strong and clearly visible.
(c) \( \mathrm{a = 2.5 \ cm = \lambda} \): The waves spread in almost all directions beyond the gap. The emerging wavefronts are circular — this is maximum diffraction.
Conclusion: Diffraction is most pronounced when the gap width is comparable to the wavelength of the wave (\( \mathrm{a \approx \lambda} \)).