What Happens When You Put a Glass Slit in YDSE?
Mastering the "Path Introduced" Concept and Fringe Shift Calculation for JEE Main & Advanced.
The Glass Slit Effect in YDSE
When a glass slab is placed in front of one slit in Young's Double Slit Experiment, it introduces an additional path difference that shifts the entire interference pattern. This is a frequently tested concept in JEE that requires understanding of:
- Optical path length vs geometrical path length
- Fringe shift calculation and direction
- Central fringe position determination
- Problem-solving approaches for different scenarios
The "Path Introduced" Concept
YDSE with Glass Slit Setup
[S2] ------------------------| |
Glass slab of thickness t and refractive index μ placed in front of S1
🔍 Additional Path Difference Introduced:
Step 1: Without glass, path from S1 = $d_1$
Step 2: With glass, optical path from S1 = $(d_1 - t) + μt$
Step 3: Additional path introduced = $[(d_1 - t) + μt] - d_1$
Step 4: Simplify: $= μt - t = t(μ - 1)$
$$\Delta x_{\text{additional}} = t(μ - 1)$$
💡 Key Insight:
The glass slab increases the optical path length from that slit by $t(μ - 1)$. This causes the entire fringe pattern to shift towards the slit with the glass slab.
Fringe Shift Calculation
📐 Fringe Shift Derivation:
Step 1: Fringe width in YDSE: $\beta = \frac{\lambda D}{d}$
Step 2: Additional path difference: $\Delta x = t(μ - 1)$
Step 3: This additional path is equivalent to shifting m fringes:
$t(μ - 1) = m\lambda$
Step 4: Number of fringes shifted:
$m = \frac{t(μ - 1)}{\lambda}$
Step 5: Actual fringe shift on screen:
$$\text{Fringe shift} = m\beta = \frac{t(μ - 1)}{\lambda} \cdot \frac{\lambda D}{d} = \frac{D}{d}t(μ - 1)$$
🎯 JEE Application Example:
Problem: Glass slab (μ=1.5, t=1.2μm) placed in front of one slit. λ=600nm. Find fringe shift if D=1m, d=1mm.
Solution: Using the formula:
$\text{Shift} = \frac{D}{d}t(μ - 1) = \frac{1}{0.001} \times 1.2\times10^{-6} \times (1.5 - 1)$
$= 1000 \times 1.2\times10^{-6} \times 0.5 = 0.6\times10^{-3} m = 0.6 mm$
Central Fringe Position
🎯 Finding New Central Maximum:
Step 1: Central fringe occurs where path difference = 0
Step 2: With glass, effective path difference:
$\Delta x_{\text{effective}} = \Delta x_{\text{geometrical}} + t(μ - 1)$
Step 3: For central fringe, set $\Delta x_{\text{effective}} = 0$:
$\Delta x_{\text{geometrical}} + t(μ - 1) = 0$
Step 4: $\Delta x_{\text{geometrical}} = -t(μ - 1)$
Step 5: Using $\Delta x_{\text{geometrical}} = \frac{yd}{D}$:
$\frac{yd}{D} = -t(μ - 1)$
$$y_{\text{central}} = -\frac{D}{d}t(μ - 1)$$
💡 Physical Interpretation:
The negative sign indicates the central fringe shifts towards the slit with glass slab. If glass is in front of left slit (S1), pattern shifts to the left (negative y-direction).
🚀 Problem-Solving Strategies
Quick Formulas:
- Additional path = $t(μ - 1)$
- Fringe shift = $\frac{D}{d}t(μ - 1)$
- Fringes shifted = $\frac{t(μ - 1)}{\lambda}$
- Central fringe: $y = -\frac{D}{d}t(μ - 1)$
JEE Exam Tips:
- Always note which slit has glass
- Remember shift direction: towards glass slit
- Watch units: convert all to meters
- Practice both numerical and conceptual questions
Advanced Variations Available
Includes glass in both slits, multiple glass slabs, inclined slabs, and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Glass (μ=1.5, t=1.8μm) in front of S1. λ=600nm. How many fringes shift?
2. If central fringe shifts by 5β, find t when μ=1.6, λ=500nm
3. Glass in both slits: μ₁=1.5, t₁=2μm; μ₂=1.6, t₂=1.5μm. Find net shift.
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