Mastering the Lens Displacement Method: Finding Focal Length Without Formula
Learn the classic two-position method for convex lenses with complete derivation and JEE-level applications.
Why the Displacement Method is a JEE Favorite
The lens displacement method (two-position method) is a classic experimental technique that appears frequently in JEE Physics because it:
- Eliminates systematic errors in focal length measurement
- Works without knowing the exact lens formula
- Provides experimental verification of lens principles
- Tests conceptual understanding beyond rote learning
The Two-Position Method: Basic Principle
Experimental Setup
Lens displacement between two sharp image positions = d
🎯 Key Insight:
For a convex lens, there are two positions where a sharp image forms on the screen when the object-screen distance is greater than 4f.
In one position, the image is magnified; in the other, it's diminished.
The focal length can be calculated using only the displacement between these two positions!
Mathematical Derivation of the Formula
Step 1: Define Variables
Let $D$ = fixed distance between object and screen
Let $d$ = distance between two lens positions giving sharp images
Let $f$ = focal length of the convex lens
Step 2: Lens Positions
In Position 1: Object distance = $u$, Image distance = $v$
From lens formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
Also, $u + v = D$ (fixed object-screen distance)
Step 3: Second Position
In Position 2: The lens is moved by distance $d$
Now, Object distance = $v$ (previous image distance)
Image distance = $u$ (previous object distance)
The roles of object and image distances are swapped!
Step 4: Solve for Focal Length
From Position 1: $u_1 = \frac{D - d}{2}$
From Position 2: $u_2 = \frac{D + d}{2}$
Using lens formula for Position 1:
$\frac{1}{f} = \frac{1}{u_1} + \frac{1}{D - u_1}$
Substitute $u_1 = \frac{D - d}{2}$:
$\frac{1}{f} = \frac{2}{D - d} + \frac{2}{D + d}$
$\frac{1}{f} = \frac{2(D + d) + 2(D - d)}{(D - d)(D + d)}$
$\frac{1}{f} = \frac{4D}{D^2 - d^2}$
$$f = \frac{D^2 - d^2}{4D}$$
JEE-Level Problem Solving
🎯 JEE Problem Example:
Problem: In a displacement method experiment with a convex lens, the distance between object and screen is 80 cm. The displacement between two positions of the lens for sharp images is 30 cm. Calculate the focal length.
Solution: Using the displacement method formula:
Given: $D = 80$ cm, $d = 30$ cm
$f = \frac{D^2 - d^2}{4D} = \frac{80^2 - 30^2}{4 \times 80}$
$f = \frac{6400 - 900}{320} = \frac{5500}{320} = 17.1875$ cm
Answer: Focal length = 17.19 cm
🔍 Experimental Verification:
Condition for two positions: $D > 4f$
In our example: $4f = 4 \times 17.19 = 68.76$ cm
$D = 80$ cm $> 68.76$ cm ✓ Condition satisfied
Magnification relation: $m_1 \times m_2 = 1$
Where $m_1$ and $m_2$ are magnifications in two positions
🚀 Problem-Solving Strategies
Key Points to Remember:
- Two sharp images form only when $D > 4f$
- Magnifications are reciprocals: $m_1 \times m_2 = 1$
- No need to measure u and v separately
- Eliminates error due to lens position
JEE Exam Tips:
- Always verify $D > 4f$ condition first
- Remember the formula: $f = \frac{D^2 - d^2}{4D}$
- Practice numerical variations
- Understand the physical significance
Advanced Applications Available
Includes error analysis, combination of lenses, and JEE Advanced level variations
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. In displacement method, D = 100 cm and d = 40 cm. Find focal length and verify if two positions exist.
2. If magnification in one position is 2, what is the magnification in the second position?
3. Prove that the condition for two positions is D > 4f.
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