Lloyd's Mirror: The Interference Experiment with a Built-In Phase Change
Understand why this experiment produces complementary fringes to YDSE and master the crucial concept of phase change on reflection.
Why Lloyd's Mirror is Special
Lloyd's Mirror demonstrates one of the most important concepts in wave optics: phase change of π radians when light reflects from a denser medium. This single phenomenon makes its interference pattern complementary to Young's Double Slit Experiment.
🎯 JEE Significance
This experiment appears frequently in JEE as it tests fundamental understanding of wave nature of light. Mastering it can help you secure 2-4 marks and understand more complex interference phenomena.
🚀 Quick Navigation
1. Experimental Setup
The Simple Yet Brilliant Design
• S: Monochromatic light source
• Two coherent sources: Direct (S) and Virtual (S') from reflection
• Interference occurs on the screen
Key Components
- Monochromatic light source (S)
- Plane mirror placed vertically
- Screen for observing interference
- Source very close to mirror plane
Working Principle
- Light reaches screen via two paths
- Direct path: Straight from source
- Reflected path: Via mirror reflection
- Two waves interfere on screen
2. The Crucial Phase Change
Reflection from Denser Medium
The Golden Rule of Phase Change
When light reflects from a denser medium, it suffers a phase change of π radians (180°)
In Lloyd's Mirror:
- Mirror acts as denser medium
- Reflected ray gets additional phase of π
- Direct ray has no phase change
- This creates inherent path difference
Mathematical Representation:
Direct wave: $y_1 = a \sin(\omega t)$
Reflected wave: $y_2 = a \sin(\omega t + \pi)$
$y_2 = -a \sin(\omega t)$
They are completely out of phase initially
Effective Path Difference
Due to the phase change of π upon reflection:
Actual path difference: $\Delta x = S'P - SP$
But effective path difference: $\Delta x_{eff} = \Delta x + \frac{\lambda}{2}$
The extra $\frac{\lambda}{2}$ comes from the π phase change
$$\Delta x_{eff} = (S'P - SP) + \frac{\lambda}{2}$$
3. Fringe Pattern Analysis
Bright and Dark Fringes
Condition for Bright Fringes
For constructive interference:
$$\Delta x_{eff} = n\lambda$$
$$(S'P - SP) + \frac{\lambda}{2} = n\lambda$$
$$S'P - SP = (n - \frac{1}{2})\lambda$$
Bright fringes when path difference = $(n - \frac{1}{2})\lambda$
Condition for Dark Fringes
For destructive interference:
$$\Delta x_{eff} = (n + \frac{1}{2})\lambda$$
$$(S'P - SP) + \frac{\lambda}{2} = (n + \frac{1}{2})\lambda$$
$$S'P - SP = n\lambda$$
Dark fringes when path difference = $n\lambda$
Key Observations
Central Fringe
- At point directly opposite to S
- Path difference = 0
- But effective path difference = $\frac{\lambda}{2}$
- Central fringe is DARK
Fringe Width
- Fringe width, $\beta = \frac{\lambda D}{d}$
- Same as YDSE formula
- D: Distance to screen
- d: Distance between S and S'
4. Lloyd's Mirror vs Young's Double Slit
Complementary Fringe Patterns
The Key Difference
Lloyd's Mirror produces exactly complementary fringes to YDSE
Where YDSE has bright fringe, Lloyd's Mirror has dark fringe, and vice versa
| Feature | Young's Double Slit | Lloyd's Mirror |
|---|---|---|
| Central Fringe | Bright | Dark |
| Condition for Bright | $\Delta x = n\lambda$ | $\Delta x = (n-\frac{1}{2})\lambda$ |
| Condition for Dark | $\Delta x = (n+\frac{1}{2})\lambda$ | $\Delta x = n\lambda$ |
| Phase Change | No inherent phase change | π phase change on reflection |
| Fringe Width | $\beta = \frac{\lambda D}{d}$ | $\beta = \frac{\lambda D}{d}$ |
💡 Memory Aid
"Lloyd Looks Different" - Remember that Lloyd's Mirror has different (complementary) fringes compared to YDSE due to the phase change on reflection.
5. Mathematical Treatment
Complete Interference Conditions
Position of Bright Fringes
For bright fringes: $S'P - SP = (n - \frac{1}{2})\lambda$
Using geometry: $\frac{x d}{D} = (n - \frac{1}{2})\lambda$
$$x = \frac{(n - \frac{1}{2})\lambda D}{d}$$
Bright fringes at $x = \frac{(2n-1)\lambda D}{2d}$
Position of Dark Fringes
For dark fringes: $S'P - SP = n\lambda$
Using geometry: $\frac{x d}{D} = n\lambda$
$$x = \frac{n\lambda D}{d}$$
Dark fringes at $x = \frac{n\lambda D}{d}$
🎯 JEE Practice Problems
Problem 1: In Lloyd's mirror experiment, the central fringe is observed to be dark. Why?
Problem 2: The fringe width in Lloyd's mirror is 0.2 mm. If wavelength is 6000 Å and screen distance is 1m, find distance between source and its virtual image.
Problem 3: Compare positions of 3rd bright fringe in YDSE and Lloyd's mirror with same setup parameters.
📋 Key Takeaways for JEE
Must Remember Facts
- Central fringe is DARK in Lloyd's Mirror
- Due to π phase change on reflection from denser medium
- Fringes are complementary to YDSE
- Fringe width formula is same as YDSE
Problem Solving Approach
- Identify the phase change factor
- Add $\frac{\lambda}{2}$ to actual path difference
- Apply standard interference conditions
- Remember complementary nature with YDSE
🎓 Conceptual Mastery
The entire difference from YDSE comes from the π phase change on reflection from denser medium.
Lloyd's Mirror provides experimental evidence for phase change on reflection.
Once you account for the extra $\frac{\lambda}{2}$, the mathematics is identical to YDSE.
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