Lens Maker's Formula Demystified: Derivation and Applications
Master the fundamental formula that governs lens behavior in different media with complete step-by-step derivation.
Why Lens Maker's Formula is Crucial for JEE
The Lens Maker's Formula is the cornerstone of geometrical optics in JEE. It connects the focal length of a lens with its physical properties and the surrounding medium, making it essential for:
🎯 JEE Importance
- 2-3 questions per paper directly use this formula
- Essential for combination of lenses problems
- Critical for understanding lens behavior in different media
- Forms basis for optical instrument calculations
🔍 Quick Navigation
1. The Lens Maker's Formula
The Standard Formula
Lens Maker's Formula
$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Where:
- $f$ = Focal length of the lens
- $\mu$ = Refractive index of lens material relative to surrounding medium
- $R_1$ = Radius of curvature of first surface
- $R_2$ = Radius of curvature of second surface
For different media:
$$\frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Where $\mu_l$ = refractive index of lens, $\mu_m$ = refractive index of medium
📐 Sign Convention (Cartesian)
For Distances:
- Distances measured along direction of incident light: Positive (+)
- Distances measured opposite to direction of incident light: Negative (-)
- Incident light direction: Left to Right
For Curvature:
- Center of curvature on same side as outgoing light: Positive (+)
- Center of curvature on opposite side to outgoing light: Negative (-)
2. Step-by-Step Derivation
Assumptions for Derivation
- Lens is thin (thickness << radii of curvature)
- Aperture is small compared to focal length
- Light rays are paraxial (make small angles with principal axis)
- Lens is placed in a homogeneous medium
Derivation for Double Convex Lens
Step 1: First Refraction at Surface 1
For refraction at spherical surface:
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1}$$
For air to glass at surface 1:
$$\frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1}$$
Step 2: Second Refraction at Surface 2
The image formed by surface 1 acts as virtual object for surface 2.
For glass to air at surface 2:
$$\frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2}$$
Note: $R_2$ is negative for convex surface facing right
Step 3: Add Both Equations
Adding equations from Step 1 and Step 2:
$$\frac{\mu}{v_1} - \frac{1}{u} + \frac{1}{v} - \frac{\mu}{v_1} = \frac{\mu - 1}{R_1} + \frac{1 - \mu}{R_2}$$
Simplifying:
$$\frac{1}{v} - \frac{1}{u} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Step 4: Lens Formula Relation
For object at infinity ($u \to \infty$), image forms at focus ($v = f$):
$$\frac{1}{f} - 0 = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Final Formula:
$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
💡 Derivation Tips for JEE
- Remember the refraction formula for spherical surfaces
- Pay attention to sign conventions for $R_1$ and $R_2$
- For concave surfaces, radius is negative
- Practice deriving for different lens types
3. Sign Convention Mastery
Radius of Curvature Signs
Convex Surface (Facing Left)
Center of curvature on right side
$R$ = Negative
Example: $R = -20$ cm
Concave Surface (Facing Left)
Center of curvature on left side
$R$ = Positive
Example: $R = +15$ cm
Common Lens Types
| Lens Type | $R_1$ | $R_2$ | $\frac{1}{R_1} - \frac{1}{R_2}$ | Nature |
|---|---|---|---|---|
| Double Convex | +$R_1$ | -$R_2$ | Positive | Converging |
| Double Concave | -$R_1$ | +$R_2$ | Negative | Diverging |
| Plano-Convex | +$R_1$ | ∞ | Positive | Converging |
| Plano-Concave | -$R_1$ | ∞ | Negative | Diverging |
4. Practical Applications
Application 1: Lens in Liquid
Modified Formula
When lens is immersed in liquid:
$$\frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Where:
- $\mu_l$ = Refractive index of lens material
- $\mu_m$ = Refractive index of surrounding medium
Example Problem
Problem: A convex lens ($\mu = 1.5$, $R_1 = 20$ cm, $R_2 = -20$ cm) is immersed in water ($\mu = 1.33$). Find its focal length.
Solution:
$$\frac{1}{f} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{20} - \frac{1}{-20}\right)$$
$$\frac{1}{f} = (1.128 - 1)\left(\frac{1}{20} + \frac{1}{20}\right)$$
$$\frac{1}{f} = 0.128 \times \frac{2}{20} = 0.0128$$
$$f = 78.125 \text{ cm}$$
Application 2: Silvered Lens (Mirror-Lens Combination)
Equivalent Focal Length
When one surface of lens is silvered, it behaves as a mirror:
$$\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}$$
Where:
- $f_l$ = focal length of lens part
- $f_m$ = focal length of mirror part
Application 3: Lens Power in Different Media
Power Calculation
Lens power changes when medium changes:
$$P = \frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Key Insight: If $\mu_l = \mu_m$, power becomes zero (lens disappears)
Application 4: Combination of Lenses
Equivalent Focal Length
For two lenses in contact:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$
Using Lens Maker's Formula for each:
$$\frac{1}{F} = (\mu_1 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) + (\mu_2 - 1)\left(\frac{1}{R_3} - \frac{1}{R_4}\right)$$
5. Practice Problems
Test Your Understanding
Problem 1: A double convex lens has radii of curvature 20 cm and 30 cm. If refractive index is 1.5, find its focal length in air.
Problem 2: The same lens from Problem 1 is immersed in water ($\mu = 1.33$). Calculate its new focal length.
Problem 3: A plano-convex lens has focal length 20 cm in air. If its refractive index is 1.5, find the radius of curvature of curved surface.
Problem 4: A convex lens of focal length 10 cm is combined with a concave lens of focal length 20 cm. Find the equivalent focal length.
📋 Quick Reference Guide
Key Formulas
- $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ (in air)
- $\frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ (in medium)
- $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ (lenses in contact)
- $P = \frac{1}{f}$ (Lens power)
Sign Convention
- Convex surface (facing left): $R$ = Negative
- Concave surface (facing left): $R$ = Positive
- Converging lens: $f$ = Positive
- Diverging lens: $f$ = Negative
Special Cases
$R_1 = +R$, $R_2 = -R$
$\frac{1}{f} = \frac{2(\mu-1)}{R}$
$R_1 = +R$, $R_2 = \infty$
$\frac{1}{f} = \frac{\mu-1}{R}$
🎯 JEE Exam Strategy
Memorize the formula and sign convention for faster problem solving.
Identify lens type quickly to apply correct sign convention.
Always check if your answer makes physical sense (converging/diverging).
Clearly indicate sign conventions used for partial credit.
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