CIE AS/A Level Physics 8.3 Interference Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 8.3 Interference Study Notes – New Syllabus
CIE AS/A Level Physics 8.3 Interference 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:
- understand the terms interference and coherence
- show an understanding of experiments that demonstrate two-source interference using water waves in a ripple tank, sound, light and microwaves
- understand the conditions required if two-source interference fringes are to be observed
- recall and use λ = ax / D for double-slit interference using light
Interference
Interference is the phenomenon that occurs when two or more waves of the same type overlap at a point in space, resulting in a new wave pattern. The resultant displacement at any point is the sum of the displacements of the individual waves at that point (as given by the principle of superposition).
Types of Interference:
Constructive Interference: Occurs when waves meet in phase (crest meets crest or trough meets trough).
\( \mathrm{Resultant \ amplitude = A_1 + A_2} \)
The intensity of the resultant wave is maximum at these points.
Destructive Interference: Occurs when waves meet in antiphase (crest meets trough).
\( \mathrm{Resultant \ amplitude = |A_1 – A_2|} \)
The waves cancel each other partially or completely, leading to a minimum in intensity.
Coherence:
Two sources are said to be coherent if they emit waves of the same frequency and maintain a constant phase difference between them over time.
Conditions for Sustained (Observable) Interference:
- The sources must emit waves of the same frequency.
- The sources must have a constant phase relationship — i.e., they must be coherent.
- The waves must have approximately the same amplitude for clear interference patterns to appear.
- The waves must be of the same type (both mechanical, both sound, or both electromagnetic).
Path Difference and Phase Difference:
\( \mathrm{Path \ difference = n\lambda} \) → Constructive interference (bright fringe or maximum)
\( \mathrm{Path \ difference = (n + \tfrac{1}{2})\lambda} \) → Destructive interference (dark fringe or minimum)
Interference patterns result from the superposition of coherent waves. The stability of these patterns depends on maintaining a fixed phase relationship between the sources — the essence of coherence.
Example
Two coherent sources emit waves of wavelength \( \mathrm{5 \ cm} \). At a certain point, the path difference between the waves from the two sources is \( \mathrm{2.5 \ cm} \). What type of interference occurs at that point?
▶️ Answer / Explanation
Given: \( \mathrm{\lambda = 5 \ cm} \), path difference \( \mathrm{= 2.5 \ cm = \tfrac{\lambda}{2}} \).
Since the path difference is half a wavelength, the waves meet in antiphase (crest meets trough), resulting in destructive interference.
Therefore, the resultant intensity is minimum at that point.
Two-Source Interference Experiments
Two-source interference occurs when waves from two coherent sources overlap and produce a pattern of alternating maxima (constructive interference) and minima (destructive interference).
The interference pattern depends on the wavelength of the wave and the geometry of the setup. The phenomenon can be demonstrated using different types of waves — water, sound, light, and microwaves.
| Type of Wave | Experimental Setup | Observation / Pattern | Conclusion |
|---|---|---|---|
| Water waves (Ripple tank) | Two identical dippers vibrate in phase in a ripple tank, producing circular waves. The overlapping wavefronts are observed on the tank floor. | Alternating regions of high and low amplitude (constructive and destructive interference) form a stationary pattern of bright and dark lines. | Water wave crests and troughs interfere to form an interference pattern — evidence of wave superposition. |
| Sound waves | Two loudspeakers connected to the same signal generator produce sound waves of the same frequency. A microphone detects sound intensity at different points. | Alternating loud (maxima) and quiet (minima) regions are heard as the microphone moves through the field. | Sound intensity varies due to interference — confirms the wave nature of sound. |
| Light waves (Young’s double-slit experiment) | Light from a single monochromatic source passes through two narrow slits that act as coherent sources. A screen displays the resulting pattern. | Bright and dark fringes appear — bright where waves meet in phase, dark where they meet in antiphase. | The pattern shows interference of light, confirming its wave nature and coherence requirement. |
| Microwaves | A microwave transmitter and two slits or reflecting surfaces act as coherent sources. A detector measures signal intensity across the region. | Alternating regions of high and low intensity are detected, corresponding to constructive and destructive interference. | Microwaves exhibit interference like light and sound, showing that all electromagnetic waves behave similarly. |
Key Conditions for Two-Source Interference:
- The sources must be coherent (same frequency, constant phase difference).
- The amplitude of both waves should be approximately equal for clear fringes.
- The path difference between the two waves determines whether the interference is constructive or destructive.
Path Difference Relationships:
Constructive interference: \( \mathrm{path \ difference = n\lambda} \)
Destructive interference: \( \mathrm{path \ difference = (n + \tfrac{1}{2})\lambda} \)
Example
In Young’s double-slit experiment, the wavelength of light is \( \mathrm{6.0 \times 10^{-7} \ m} \), and the distance between the slits is \( \mathrm{0.3 \ mm} \). The screen is \( \mathrm{1.2 \ m} \) away. Calculate the distance between adjacent bright fringes (fringe spacing).
▶️ Answer / Explanation
The fringe spacing \( \mathrm{\Delta y} \) is given by:
\( \mathrm{\Delta y = \dfrac{\lambda D}{a}} \)
Substituting: \( \mathrm{\Delta y = \dfrac{6.0 \times 10^{-7} \times 1.2}{0.3 \times 10^{-3}} = 2.4 \times 10^{-3} \ m = 2.4 \ mm} \)
Therefore, the adjacent bright fringes are spaced 2.4 mm apart on the screen.
Conditions for Observing Two-Source Interference Fringes
Two-source interference fringes are the alternating bright and dark bands formed due to the constructive and destructive interference of waves from two coherent sources. To observe a clear and stable interference pattern, specific conditions must be satisfied.
Essential Conditions:
- 1. The two sources must be coherent.
They must emit waves of the same frequency and maintain a constant phase difference. Without coherence, the interference pattern fluctuates and becomes invisible. - 2. The two sources must have the same wavelength (monochromatic waves).
If the sources emit multiple wavelengths (like white light), the fringes from different colors overlap and blur the pattern. - 3. The amplitudes of the two waves should be approximately equal.
Unequal amplitudes cause incomplete cancellation during destructive interference, making dark fringes less distinct. - 4. The waves must overlap in space.
Interference occurs only in regions where both waves meet and superpose. - 5. The path difference between the two waves should be small compared to the distance to the screen.
This ensures that fringes are evenly spaced and visible. - 6. The experimental setup should be stable and vibration-free.
External disturbances can destroy the steady phase relationship required for interference.
Additional Note for Light Interference:
- In light experiments (e.g., Young’s double-slit), both slits are usually illuminated by the same single source to ensure coherence.
- For white light, only a few colored fringes (centered with white) are visible before the pattern fades due to overlapping wavelengths.
Example
Explain why two ordinary electric bulbs do not produce a visible interference pattern on a screen when their light overlaps.
▶️ Answer / Explanation
Ordinary bulbs emit light waves that are incoherent — their wavefronts have random phase differences and vary continuously. As a result, the phase relationship between the two sources changes rapidly, and any interference fringes average out, leaving uniform illumination.
Therefore, coherent sources are necessary for observable interference fringes.
Double-Slit Interference Formula — \( \mathrm{\lambda = \dfrac{a x}{D}} \)
In Young’s double-slit experiment, monochromatic light passes through two narrow slits separated by distance \( \mathrm{a} \). The slits act as coherent sources, and the resulting light waves interfere on a distant screen placed at distance \( \mathrm{D} \) from the slits, forming equally spaced bright and dark fringes.
Expression for Fringe Spacing:
\( \mathrm{x = \dfrac{\lambda D}{a}} \)
where:
- \( \mathrm{x} \): distance between adjacent bright (or dark) fringes (fringe spacing)
- \( \mathrm{\lambda} \): wavelength of the light used
- \( \mathrm{a} \): separation between the two slits
- \( \mathrm{D} \): distance between the double slits and the screen
Rearranging:
\( \mathrm{\lambda = \dfrac{a x}{D}} \)
DerivationF:
- At a point on the screen, constructive interference occurs when the path difference between the two waves is an integer multiple of the wavelength: \( \mathrm{a \sin \theta = n\lambda} \).
- For small angles (\( \mathrm{\theta} \) is small), \( \mathrm{\sin \theta \approx \tan \theta = \dfrac{x}{D}} \).
- Hence, \( \mathrm{a \dfrac{x}{D} = n\lambda} \Rightarrow \dfrac{x}{n} = \dfrac{\lambda D}{a}} \).
- So, the distance between adjacent fringes (for successive \( \mathrm{n} \)) is \( \mathrm{x = \dfrac{\lambda D}{a}} \).
Key Relationships:
- Fringe spacing increases with wavelength (λ) and screen distance (D).
- Fringe spacing decreases as slit separation (a) increases.
Example
In a double-slit experiment, light of wavelength \( \mathrm{5.0 \times 10^{-7} \ m} \) passes through slits separated by \( \mathrm{0.25 \ mm} \). The screen is \( \mathrm{1.5 \ m} \) away. Calculate the fringe spacing.
▶️ Answer / Explanation
Given: \( \mathrm{\lambda = 5.0 \times 10^{-7} \ m, \ a = 0.25 \times 10^{-3} \ m, \ D = 1.5 \ m.} \)
Using \( \mathrm{x = \dfrac{\lambda D}{a}} \):
\( \mathrm{x = \dfrac{5.0 \times 10^{-7} \times 1.5}{0.25 \times 10^{-3}} = 3.0 \times 10^{-3} \ m = 3.0 \ mm.} \)
Therefore, the distance between adjacent bright fringes is \( \mathrm{3.0 \ mm.} \)