Prerequisites
Before studying this topic, make sure you understand:
- Reflection and Refraction — Basic lens formula and sign conventions
- Snell’s law and refractive index
- Basic geometry and trigonometry
The Hook: The Superhero Power of Focus
Ever notice how Iron Man’s arc reactor uses lenses to focus light into a beam? Or how in Spider-Man: Far From Home, Mysterio uses holograms and lenses to create illusions? Lenses control where light goes — and that control is power!
Why can some people see far but not near? Why do cameras zoom? Why does a magnifying glass burn paper? All because lenses bend light differently based on their shape and material. Let’s crack the code!
JEE Weightage: HIGH — Expect 1-2 questions in JEE Main, 1 in Advanced (especially lens combinations)
The Core Concept
A thin lens is a piece of transparent material (usually glass) with two curved surfaces, thin enough that we can ignore the thickness in calculations.
Types of Lenses
| Type | Shape | Effect | Focal Length |
|---|---|---|---|
| Convex (Converging) | Thicker in middle | Converges parallel rays | Positive |
| Concave (Diverging) | Thinner in middle | Diverges parallel rays | Negative |
In simple terms: Convex lenses bring light together (like a magnifying glass), concave lenses spread light apart (like prescription glasses for nearsightedness).
Ray Diagram for Convex Lens
Lens Maker’s Equation
This is THE formula that tells you what focal length a lens will have based on its shape and material.
$$\boxed{\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$Where:
- $f$ = focal length of the lens
- $n$ = 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
Sign Convention for Radii
Measure radii from the surface:
- If center of curvature is on the right (image side): $R$ is positive
- If center of curvature is on the left (object side): $R$ is negative
Mnemonic: “Right is Right” (Right side = positive R)
Common Lens Shapes
Double Convex (Biconvex):
- $R_1 > 0$ (first surface curves right)
- $R_2 < 0$ (second surface curves right, but we measure from second surface, so negative)
- Both surfaces converge light → strong positive lens
Plano-Convex:
- $R_1 > 0$ (curved surface)
- $R_2 = \infty$ (plane surface)
- $\frac{1}{f} = (n-1) \cdot \frac{1}{R_1}$
Double Concave (Biconcave):
- $R_1 < 0$
- $R_2 > 0$
- Both surfaces diverge light → strong negative lens
Power of a Lens
The power of a lens measures its ability to converge or diverge light.
$$\boxed{P = \frac{1}{f}}$$Units:
- $f$ in meters → $P$ in diopters (D)
- 1 D = 1 m$^{-1}$
In simple terms: Higher power = stronger bending of light. Short focal length = high power!
Properties of Power
- Convex lens: $P > 0$ (positive power)
- Concave lens: $P < 0$ (negative power)
- Short focal length = High power (strong lens)
- Long focal length = Low power (weak lens)
Example:
- Reading glasses: $P = +2.5$ D → $f = \frac{1}{2.5} = 0.4$ m = 40 cm
- Nearsightedness glasses: $P = -1.5$ D → $f = -\frac{1}{1.5} = -0.67$ m
Combination of Lenses
When two or more thin lenses are placed in contact, their powers add up!
Lenses in Contact
$$\boxed{P_{total} = P_1 + P_2 + P_3 + ...}$$Or equivalently:
$$\boxed{\frac{1}{f_{total}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + ...}$$In simple terms: Powers add like resistances in parallel!
Memory Trick: “Powers Pile up Perfectly” → $P_{total} = P_1 + P_2$
Example
Two lenses in contact: $f_1 = 20$ cm, $f_2 = -30$ cm
$$\frac{1}{f} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60}$$ $$f = 60 \text{ cm}$$The combination acts like a single convex lens of focal length 60 cm!
Lenses Separated by Distance
When lenses are separated by distance $d$:
$$\boxed{\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}}$$Special Case: If $d = 0$ (in contact), this reduces to $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
For separated lenses, you CANNOT simply add powers! You must either:
- Use the formula above, OR
- Trace the image step-by-step (image by first lens becomes object for second lens)
For JEE Advanced, always use the step-by-step method for clarity.
Memory Tricks & Patterns
Mnemonic for Lens Maker’s Equation
Memory Trick: “Nisha Really Runs” → $(n-1)(\frac{1}{R_1} - \frac{1}{R_2})$
Mnemonic for Power Addition
Memory Trick: “Pile Powers” → $P = P_1 + P_2$
Pattern Recognition
- Convex + Convex = Stronger convex (powers add)
- Concave + Concave = Stronger concave (powers add in negative)
- Convex + Concave = Depends on which is stronger
- If $|P_{convex}| > |P_{concave}|$ → Resultant convex
- If $|P_{concave}| > |P_{convex}|$ → Resultant concave
Quick Decision Tree
Finding focal length?
├─ Given R₁, R₂, n? → Use Lens Maker's Equation
└─ Given object/image positions? → Use Lens Formula
Lens combination?
├─ In contact? → Add powers: P = P₁ + P₂
└─ Separated? → Use step-by-step image tracing
Lens in Different Media
If a lens (refractive index $n_L$) is placed in a medium (refractive index $n_m$):
$$\boxed{\frac{1}{f} = \left(\frac{n_L}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$Special Cases
Lens in water: If $n_{glass} = 1.5$ and $n_{water} = 1.33$:
$$\frac{1}{f_{water}} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$The effective power decreases in water!
Extreme case: If $n_L = n_m$ (lens material = medium), then $f = \infty$ → lens disappears optically!
Common Mistakes to Avoid
Wrong: Using $R_1 = 20$ cm, $R_2 = 30$ cm for a double convex lens
Correct: For double convex, second surface curves the other way:
- $R_1 = +20$ cm
- $R_2 = -30$ cm
Always draw the lens and mark which side the centers are!
Wrong: $P_{total} = P_1 + P_2$ for lenses with separation
Correct: Use $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$ OR trace images step-by-step
JEE Loves This Trap! Read carefully — “in contact” or “separated by”?
Wrong: Using absolute refractive index when lens is in water
Correct: Use relative refractive index $\frac{n_{lens}}{n_{medium}}$
The surrounding medium matters!
Wrong: $f = 20$ cm → $P = \frac{1}{20} = 0.05$ D
Correct: Convert to meters first! $f = 0.2$ m → $P = \frac{1}{0.2} = 5$ D
Always convert focal length to meters before calculating power!
Practice Problems
Level 1: Foundation (NCERT-type)
A double convex lens has radii of curvature 10 cm and 15 cm. The refractive index of glass is 1.5. Find the focal length.
Solution: For double convex:
- $R_1 = +10$ cm (first surface, center on right)
- $R_2 = -15$ cm (second surface, center on left when measured from second surface)
- $n = 1.5$
Answer: 12 cm (converging lens)
A lens has a focal length of 25 cm. Find its power.
Solution: Convert to meters: $f = 0.25$ m
$$P = \frac{1}{f} = \frac{1}{0.25} = 4 \text{ D}$$Answer: +4 D (converging lens)
Two lenses of focal lengths 20 cm and 30 cm are placed in contact. Find the focal length of the combination.
Solution:
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20} + \frac{1}{30}$$ $$\frac{1}{f} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$$ $$f = 12 \text{ cm}$$Answer: 12 cm
Level 2: JEE Main
A plano-convex lens ($n = 1.5$) has a curved surface with radius 20 cm. What is its focal length? If the plane surface is now silvered (converted to mirror), what is the effective focal length of the system?
Solution: Part 1: Just the lens
- $R_1 = 20$ cm (curved surface)
- $R_2 = \infty$ (plane surface)
Part 2: Plane surface silvered This acts as a mirror with $f_{mirror} = \infty$ (plane mirror) combined with the lens.
For lens + mirror combination: $P_{total} = P_{lens} + P_{mirror} + P_{lens}$ (light passes through lens twice)
Actually, for a silvered lens:
$$\frac{1}{f_{eq}} = \frac{1}{f_{lens}} + \frac{1}{f_{mirror}} + \frac{1}{f_{lens}} = \frac{2}{f_{lens}} + \frac{1}{f_{mirror}}$$For plane mirror: $f_{mirror} = \infty$
$$\frac{1}{f_{eq}} = \frac{2}{40} + 0 = \frac{1}{20}$$ $$f_{eq} = 20 \text{ cm}$$Answer: Lens alone: 40 cm; Silvered: 20 cm (acts like a concave mirror!)
Two thin lenses of powers +5 D and $P$ D are placed in contact. If the combination has power +3 D, find $P$.
Solution:
$$P_{total} = P_1 + P_2$$ $$3 = 5 + P$$ $$P = -2 \text{ D}$$Answer: -2 D (a concave lens is needed to reduce the power)
Two convex lenses of focal length 10 cm each are separated by 30 cm. Find the effective focal length.
Solution: Using $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$:
$$\frac{1}{f} = \frac{1}{10} + \frac{1}{10} - \frac{30}{10 \times 10}$$ $$\frac{1}{f} = 0.1 + 0.1 - 0.3 = -0.1$$ $$f = -10 \text{ cm}$$Interesting! Two convex lenses, when separated by $d = f_1 + f_2$, act like a diverging system!
Answer: -10 cm (diverging)
Level 3: JEE Advanced
A convex lens made of glass ($n = 1.5$) has a focal length of 20 cm in air. What is its focal length when immersed in water ($n = 1.33$)?
Solution: In air: $\frac{1}{f_{air}} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
So $\frac{1}{R_1} - \frac{1}{R_2} = \frac{2}{f_{air}} = \frac{2}{20} = \frac{1}{10}$
In water: $\frac{1}{f_{water}} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
$$\frac{1}{f_{water}} = (1.128 - 1) \times \frac{1}{10} = 0.128 \times 0.1 = 0.0128$$ $$f_{water} = \frac{1}{0.0128} = 78.1 \text{ cm}$$Alternative formula:
$$\frac{f_{water}}{f_{air}} = \frac{(n_{glass} - n_{air})}{(n_{glass} - n_{water})} \times \frac{n_{water}}{n_{air}}$$ $$\frac{f_{water}}{20} = \frac{1.5 - 1}{1.5 - 1.33} \times \frac{1.33}{1} = \frac{0.5}{0.17} \times 1.33 = 3.91$$ $$f_{water} = 78.2 \text{ cm}$$Answer: ~78 cm (power decreases significantly in water!)
Two lenses of focal lengths $f_1 = 10$ cm and $f_2 = 20$ cm are kept in contact. An object is placed at 20 cm from the combination. Find the position and magnification of the final image.
Solution: Step 1: Find combination focal length
$$\frac{1}{f} = \frac{1}{10} + \frac{1}{20} = \frac{2+1}{20} = \frac{3}{20}$$ $$f = \frac{20}{3} = 6.67 \text{ cm}$$Step 2: Find image position $u = -20$ cm
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$\frac{3}{20} = \frac{1}{v} - \frac{1}{-20}$$ $$\frac{1}{v} = \frac{3}{20} - \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$$ $$v = 10 \text{ cm}$$Step 3: Magnification
$$m = \frac{v}{u} = \frac{10}{-20} = -0.5$$Answer: Image at 10 cm on opposite side, inverted, half the size
A converging lens of focal length 20 cm is in contact with a diverging lens. If the power of the combination can vary from +1 D to +5 D, find the range of focal length of the diverging lens.
Solution: For converging lens: $f_1 = 20$ cm $= 0.2$ m → $P_1 = +5$ D
For combination: $P_{total} = P_1 + P_2 = 5 + P_2$
When $P_{total} = +1$ D:
$$1 = 5 + P_2 \implies P_2 = -4 \text{ D}$$ $$f_2 = -\frac{1}{4} = -0.25 \text{ m} = -25 \text{ cm}$$When $P_{total} = +5$ D:
$$5 = 5 + P_2 \implies P_2 = 0$$ $$f_2 = \infty$$Answer: Diverging lens focal length ranges from -25 cm to $-\infty$ (or from -4 D to 0 D)
Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Finding focal length from shape | $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$ |
| Double convex | Both surfaces converge → strong positive |
| Double concave | Both surfaces diverge → strong negative |
| Plano-convex | One curved, one flat → moderate power |
| Power from focal length | $P = \frac{1}{f}$ (f in meters) |
| Lenses in contact | $P = P_1 + P_2$ or $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$ |
| Lenses separated | Use step-by-step OR $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1f_2}$ |
| Lens in different medium | Use $n_{relative} = \frac{n_{lens}}{n_{medium}}$ |
| Silvered lens | $\frac{1}{f_{eq}} = \frac{2}{f_{lens}} + \frac{1}{f_{mirror}}$ |
JEE Exam Strategy
- For lens combinations in contact: Just add powers (fastest method!)
- For separated lenses: Always trace step-by-step (clearer, less error-prone)
- Check units: Convert cm to m before calculating power (saves silly mistakes)
- Sign check: Draw a quick lens diagram to verify $R_1$ and $R_2$ signs
- Silvered lens? Remember light passes through lens TWICE → double power
Common JEE Patterns:
- Lens maker’s equation with one radius infinite (plano-convex/concave)
- Lens in water → power reduces
- Combination with one lens variable → find range/specific value
- Silvered lens → equivalent mirror system
High-Yield Concepts:
- Power addition for contact lenses (easiest marks!)
- Lens in different media (tricky but formulaic)
- Separated lens systems (needs careful calculation)
Related Topics
Within Optics
- Reflection and Refraction — Foundation of lens formula
- Prism — Refraction through non-spherical surfaces
- Wave Optics — Interference effects in lenses
- Diffraction — Limits of lens resolution
Connected Chapters
- Ray Optics — Geometric construction of images
- Human Eye — Natural lens system
- Optical Instruments — Microscope, telescope designs
Math Connections
- Series and Reciprocals — Understanding $\frac{1}{f}$ additions
- Coordinate Geometry — Sign conventions
- Optimization — Finding max/min power
Teacher’s Summary
- Lens maker’s equation connects shape (R₁, R₂) and material (n) to focal length
- Power = 1/f — short focal length means high power (strong lens)
- Powers add for lenses in contact — this is the fastest method for combinations
- Medium matters — same lens has different focal length in water vs air
- Sign conventions for radii: Draw the lens, mark centers, assign signs
- Silvered lens = lens + mirror, light passes through lens twice
“Shape and material make the lens; power tells you its strength. Master the signs, and lens problems become your easiest points!”
JEE Success Mantra: Practice 15 problems on lens combinations — these are high-probability, high-scoring questions that take just 2 minutes if you know the formulas!