Thin Lens Formula - The Heart of Optics
The Movie Hook 🎬
Ever wonder how the camera lens in Interstellar captured those stunning black hole images? Or how Spider-Man’s camera autofocuses? Or why Superman’s glasses disguise his identity (spoiler: they don’t make sense!)? All camera lenses, eyeglasses, and telescopes use the lens formula we’re about to master!
The Big Picture
Lenses are transparent materials with curved surfaces that refract light to form images. Understanding lenses is crucial for:
- Cameras & Photography (focal length, aperture)
- Human eye (how we see!)
- Microscopes & Telescopes (magnification)
- Eyeglasses (correcting vision)
Connection Alert: Builds on Refraction, connects to Optical Instruments.
Core Concepts
1. Types of Lenses
Convex Lens (Converging)
- Thicker at center, thinner at edges
- Converges parallel rays to a focal point
- Positive focal length
- Can form real or virtual images
Concave Lens (Diverging)
- Thinner at center, thicker at edges
- Diverges parallel rays
- Negative focal length
- Always forms virtual, erect, diminished images
Memory Trick: “Convex Converges, Concave Cannot” (concave cannot converge)
2. Sign Convention (NEW CARTESIAN)
CRITICAL: All JEE problems use this convention!
Reference: Optical center of lens
- Left side (object side): Negative
- Right side (image side): Positive
- Above axis: Positive height
- Below axis: Negative height
Standard setup:
- Object on left → u is negative
- Light travels left to right
3. The Lens Formula
$$\boxed{\frac{1}{v} - \frac{1}{u} = \frac{1}{f}}$$Where:
- v = image distance from lens
- u = object distance from lens (negative for real objects)
- f = focal length
- Convex: f > 0 (positive)
- Concave: f < 0 (negative)
Key Difference from Mirror Formula:
- Mirror: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ (plus sign)
- Lens: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ (minus sign)
Memory Trick: “Lens uses Minus” (LM), “Mirror uses Plus” (MP backward!)
4. Lens Maker’s Equation
$$\boxed{\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$Where:
- n = refractive index of lens material (w.r.t. surrounding medium)
- R₁ = radius of curvature of first surface
- R₂ = radius of curvature of second surface
Sign Convention for R:
- Center of curvature on right of surface → R positive
- Center of curvature on left of surface → R negative
For symmetric lens: R₁ = +R, R₂ = -R
$$\frac{1}{f} = (n-1)\frac{2}{R}$$5. Power of Lens
$$\boxed{P = \frac{1}{f}}$$Where:
- P = power in diopters (D)
- f = focal length in meters
Properties:
- Convex lens: P > 0 (positive power)
- Concave lens: P < 0 (negative power)
- For combination: $$P_{total} = P_1 + P_2 + P_3 + ...$$
Memory Trick: “Power = 1 over F”
Example: f = 25 cm = 0.25 m → P = 1/0.25 = 4 D
6. Magnification
Linear Magnification:
$$\boxed{m = \frac{v}{u} = \frac{h_i}{h_o}}$$Note: Different from mirrors! (No negative sign for lenses)
Interpretation:
- m > 0: Erect image (virtual or on same side)
- m < 0: Inverted image (real)
- |m| > 1: Magnified
- |m| < 1: Diminished
- |m| = 1: Same size
7. Ray Diagram Rules
For Convex Lens:
- Ray parallel to axis → passes through F on other side
- Ray through optical center → goes straight (undeviated)
- Ray through F on object side → emerges parallel to axis
For Concave Lens:
- Ray parallel to axis → appears to come from F on same side
- Ray through optical center → goes straight
- Ray towards F on far side → emerges parallel to axis
8. Image Formation - Convex Lens
| Object Position | Image Position | Nature | Size | Magnification |
|---|---|---|---|---|
| At infinity | At F | Real, inverted | Point | m ≈ 0 |
| Beyond 2F | Between F and 2F | Real, inverted | Diminished | 0 < |
| At 2F | At 2F | Real, inverted | Same size | |
| Between F and 2F | Beyond 2F | Real, inverted | Magnified | |
| At F | At infinity | Real, inverted | Highly magnified | m → ∞ |
| Between F and O | Same side as object | Virtual, erect | Magnified | m > 1 |
Memory Trick: “Infinity Brings Focus, F Brings Infinity” (IBF, FBI)
9. Image Formation - Concave Lens
Always:
- Virtual image
- Erect
- Diminished (m < 1)
- On same side as object
Since both f and u are negative, v is always negative (same side).
The Formula Cheat Sheet
LENS ESSENTIALS:
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1. Lens formula: 1/v - 1/u = 1/f
2. Lens maker: 1/f = (n-1)(1/R₁ - 1/R₂)
3. Magnification: m = v/u = h_i/h_o
4. Power: P = 1/f (in meters)
5. Power combination: P = P₁ + P₂ + P₃...
6. In medium: 1/f_med = (n_L/n_M - 1)(1/R₁ - 1/R₂)
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Advanced Formulas
1. Lens in Different Medium
$$\boxed{\frac{1}{f_{medium}} = \left(\frac{n_{lens}}{n_{medium}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$Special Case: If lens is in water:
$$\frac{f_{water}}{f_{air}} = \frac{n_{water}(n_{lens} - 1)}{n_{lens} - n_{water}}$$2. Combination of Thin Lenses in Contact
$$\boxed{\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}}$$Or:
$$\boxed{P = P_1 + P_2}$$Net magnification:
$$m = m_1 \times m_2$$3. Two Lenses Separated by Distance d
$$\boxed{\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}}$$When d = 0: Reduces to contact formula
4. Newton’s Formula (Alternative)
$$\boxed{xy = f^2}$$Where:
- x = distance of object from first focal point
- y = distance of image from second focal point
Useful when positions are given w.r.t. focal points instead of lens.
Common Traps & Mistakes
Trap #1: Sign Convention Confusion
❌ Wrong: Using mirror formula for lens (1/v + 1/u = 1/f) ✅ Right: Lens formula has MINUS: 1/v - 1/u = 1/f
Trap #2: Magnification Sign
❌ Wrong: Using m = -v/u (mirror formula) ✅ Right: For lens: m = +v/u (no negative sign!)
Trap #3: Power Units
❌ Wrong: Using focal length in cm directly in P = 1/f ✅ Right: Convert f to meters first!
Trap #4: Lens Maker’s Equation Signs
❌ Wrong: Ignoring sign convention for R₁ and R₂ ✅ Right: Carefully apply signs based on center of curvature position
Trap #5: Combination Formulas
❌ Wrong: Using P = P₁ + P₂ for separated lenses ✅ Right: Use full formula with separation distance d
Practice Problems
Level 1: JEE Main Warmup
Problem 1.1: A convex lens has focal length 20 cm. An object is placed 30 cm from the lens. Find image position and magnification.
Solution
Given: f = +20 cm (convex), u = -30 cm
Using lens formula:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30}$$ $$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$$ $$v = 60 \text{ cm}$$Magnification:
$$m = \frac{v}{u} = \frac{60}{-30} = -2$$Answers:
- Image at 60 cm on other side (real)
- Magnification = -2 (inverted, twice the size)
Problem 1.2: A concave lens has focal length 15 cm. Find its power.
Solution
f = -15 cm = -0.15 m (negative for concave)
$$P = \frac{1}{f} = \frac{1}{-0.15} = -6.67 \text{ D}$$Answer: Power = -6.67 D (negative as expected for concave lens)
Problem 1.3: Two thin lenses of power +3 D and -2 D are placed in contact. Find the focal length of the combination.
Solution
Answer: Focal length = 100 cm (convex combination)
Level 2: JEE Main/Advanced
Problem 2.1: A symmetric convex lens (n=1.5) has radii of curvature 20 cm. Find its focal length in air and power.
Solution
Symmetric lens: R₁ = +20 cm, R₂ = -20 cm (by sign convention)
Using lens maker’s equation:
$$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{f} = (1.5-1)\left(\frac{1}{20} - \frac{1}{-20}\right)$$ $$\frac{1}{f} = 0.5 \times \left(\frac{1}{20} + \frac{1}{20}\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}$$ $$f = 20 \text{ cm} = 0.2 \text{ m}$$Power:
$$P = \frac{1}{0.2} = 5 \text{ D}$$Answers: f = 20 cm, P = 5 D
Problem 2.2: A convex lens forms a real image 4 times the size of the object. If the distance between object and image is 75 cm, find focal length.
Solution
Given: m = -4 (real image, so negative), distance between object and image = 75 cm
From magnification:
$$m = \frac{v}{u} = -4$$So:
$$v = -4u$$Since image is real and on opposite side: v is positive, u is negative. Actually:
$$v = -4u$$means if u = -x, then v = 4x
Distance:
$$v - u = 75$$ $$4x - (-x) = 75$$ $$5x = 75$$ $$x = 15$$So: u = -15 cm, v = 60 cm
Using lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{60} - \frac{1}{-15} = \frac{1}{60} + \frac{1}{15}$$ $$\frac{1}{f} = \frac{1 + 4}{60} = \frac{5}{60} = \frac{1}{12}$$ $$f = 12 \text{ cm}$$Answer: f = 12 cm
Problem 2.3: A glass lens (n=1.5) has focal length 20 cm in air. What will be its focal length when immersed in water (n=1.33)?
Solution
Using the medium formula:
$$\frac{f_{water}}{f_{air}} = \frac{n_{water}(n_{lens} - 1)}{n_{lens} - n_{water}}$$ $$\frac{f_{water}}{20} = \frac{1.33(1.5 - 1)}{1.5 - 1.33} = \frac{1.33 \times 0.5}{0.17} = \frac{0.665}{0.17} = 3.91$$ $$f_{water} = 20 \times 3.91 = 78.2 \text{ cm}$$Answer: 78.2 cm (focal length increases in denser medium!)
Insight: Lens becomes weaker in water because relative refractive index decreases.
Level 3: JEE Advanced
Problem 3.1: Two thin lenses of focal lengths f₁ = 10 cm and f₂ = 20 cm are separated by 5 cm. Find the focal length and power of the combination.
Solution
Given: f₁ = 10 cm, f₂ = 20 cm, d = 5 cm
For separated lenses:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$ $$\frac{1}{F} = \frac{1}{10} + \frac{1}{20} - \frac{5}{10 \times 20}$$ $$\frac{1}{F} = \frac{2 + 1}{20} - \frac{5}{200} = \frac{3}{20} - \frac{1}{40} = \frac{6 - 1}{40} = \frac{5}{40} = \frac{1}{8}$$ $$F = 8 \text{ cm} = 0.08 \text{ m}$$Power:
$$P = \frac{1}{0.08} = 12.5 \text{ D}$$Answers: F = 8 cm, P = 12.5 D
Note: Compare with contact (d=0):
$$\frac{1}{F} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$$→ F = 6.67 cm Separation increases focal length!
Problem 3.2: A point object is placed at distance u from a convex lens of focal length f. The lens is cut into two halves along the optical axis and the two halves are separated by distance d. Find the distance between the two images formed.
Solution
When lens is cut along optical axis, each half still has focal length f (not f/2!).
Each half acts as a separate lens with same f.
For upper half (shifted up by d/2):
- Object distance: u (same)
- Image distance: v (from lens formula, same)
- But lens is shifted vertically by d/2
For lower half (shifted down by d/2):
- Same image distance v
- But lens is shifted vertically by -d/2
Both form images at same distance v from their respective centers.
Vertical separation of images:
For an object at distance u from lens, magnification m = v/u
The object has effectively height h = 0 (point object at axis), but the shift matters.
Actually, due to vertical shift of lens halves: Image shift = m × (shift in lens) = (v/u) × d
Total separation = 2 × (v/u) × (d/2) = vd/u
But using lens formula: v = fu/(u-f)
Distance between images =
$$\frac{vd}{u} = \frac{fd}{u-f}$$Answer:
$$\Delta y = \frac{fd}{u-f}$$Alternative interpretation: If cut perpendicular to axis, each half has f’ = 2f. Then two images at different distances along axis.
Problem 3.3: A thin convex lens of focal length f is placed on a plane mirror. An object is placed at distance d above the lens. Find the position where image coincides with object.
Solution
For image to coincide with object, light must retrace its path.
This happens when object is at center of curvature of the lens-mirror system.
Step 1: Light from object passes through lens
Image formed by lens alone at distance v₁:
$$\frac{1}{v_1} - \frac{1}{-d} = \frac{1}{f}$$ $$\frac{1}{v_1} = \frac{1}{f} - \frac{1}{d}$$Step 2: This image acts as object for mirror (plane mirror)
Mirror forms image at equal distance behind it.
Step 3: This image acts as object for lens (going back)
For image to coincide with original object, the rays must be perpendicular to lens (parallel to axis after lens).
This happens when object is at focal point.
Condition: d = f
Or more precisely, when rays converge at infinity after first pass through lens, they come back parallel and converge at f after reflection.
Answer: Object should be placed at distance f (one focal length) above the lens.
Better approach: For exact coincidence: The image formed by lens should be at mirror surface, so v₁ = 0 (at mirror).
$$\frac{1}{0} - \frac{1}{-d} = \frac{1}{f}$$This gives ∞, so not correct.
Correct condition: Light should fall perpendicular to mirror, which means after lens, light should be parallel to axis.
This happens when object is at focal plane: d = f
Final Answer: d = f
Quick Revision Cards
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║ LENS QUICK FACTS ║
╚══════════════════════════════════════╝
🎯 Lens formula: 1/v - 1/u = 1/f
🎯 Convex: f > 0, Concave: f < 0
🎯 m = v/u (NO negative sign!)
🎯 P = 1/f (in meters) in Diopters
🎯 Combination: P = P₁ + P₂
🎯 Lens maker: 1/f = (n-1)(1/R₁-1/R₂)
🎯 In medium: n_rel = n_L/n_M
Real-World Applications
1. Camera Lens
- Multiple lenses for aberration correction
- f-number = focal length / aperture diameter
- Smaller f-number → more light → shallow depth of field
2. Eyeglasses
- Myopia (nearsighted): Concave lens (negative power)
- Hypermetropia (farsighted): Convex lens (positive power)
- Presbyopia: Bifocal (two different powers)
3. Magnifying Glass
- Object placed between F and lens
- Creates virtual, erect, magnified image
- Angular magnification = 25/f (f in cm)
Cross-Topic Connections
- Refraction: Lenses use double refraction
- Optical Instruments: Microscope, telescope use lenses
- Reflection: Lens-mirror combinations
- Prism: Dispersion limits lens quality (chromatic aberration)
- Wave Optics: Diffraction limits resolution
Exam Strategy
Time Allocation:
- Simple lens formula: 2 min
- Lens maker’s equation: 3 min
- Combination problems: 4 min
- Advanced combinations: 5-6 min
Common JEE Patterns:
- Lens in different media
- Multiple lens combinations
- Lens + mirror systems
- Displacement method (Bessel’s method)
- Moving object/lens problems (differentiation)
Quick Check:
- Convex lens with close object → virtual image (erect, magnified)
- Convex lens with far object → real image (inverted)
- Concave lens → always virtual, erect, diminished
Final Tips
- ALWAYS write sign convention at the start
- Remember: Lens formula has MINUS (1/v - 1/u = 1/f)
- Power problems: Convert to meters first!
- Draw ray diagrams for verification
- Combination: P = P₁ + P₂ only for contact lenses
Pro Tip: If problem says “lens in contact with mirror,” it’s likely testing the equivalent focal length: 1/f_eq = 2/f_lens + 1/f_mirror
Next up: Prism Optics - dispersion and deviation!