Thin Lens Combination
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Direct answer
Two thin lenses acting together behave as one whose power depends on their separation. In contact, the powers simply add: 1/F = 1/f1 + 1/f2. Separated by distance d, the equivalent focal length obeys 1/F = 1/f1 + 1/f2 − d/(f1f2); at zero separation this reduces to the contact case, and increasing d weakens a converging pair — the physics behind zoom systems and telephoto constructions. The total magnification is the product of the individual magnifications, m = m1 × m2, with the first lens's image serving as the second's object. The laboratory partner of these formulae is the displacement method: with object and screen a fixed distance D apart, a lens forms a sharp image at two positions separated by d, and f = (D² − d²)/(4D) — a focal length measured without ever locating the focal point.
What you must remember
- Contact combination: P = P1 + P2 in dioptres; a +5 D and a −2 D lens in contact act as +3 D, the eyewear principle behind bifocal-style stacking and achromatic pairs.
- Separated combination: 1/F = 1/f1 + 1/f2 − d/(f1f2); the subtraction means separation reduces the converging power of a convex pair and can even flip the net sign for large d.
- Magnification: m = m1m2, computed stage by stage — image from lens 1 becomes the object for lens 2, virtual or real with signs tracked.
- Displacement method: f = (D² − d²)/(4D), where D is the fixed object–screen distance and d the separation of the two conjugate lens positions; it requires D ≥ 4f, and equals it in the marginal case d = 0.
- The 4f condition: an object–screen distance below 4f admits no real-image position at all (the quadratic in lens position has no real roots) — a fact JEE tests as a conceptual one-liner.
- Practical note: equivalent focal length measured from which plane — the two-lens F is referred to the principal plane of the combination, not to the midpoint; JEE sidesteps this by thin-lens symmetry, but Advanced problems hint at it.
Two positions of one lens
Fix a luminous object and a screen D = 90 cm apart. Slide a convex lens along the rail: at one position the screen holds a sharp enlarged image; move the lens toward the screen and at a second position, d = 30 cm away, a sharp diminished image appears. The formula hands over the focal length immediately: f = (8100 − 900)/360 = 7200/360 = 20 cm. The elegance is metrological: no focal point, no infinity adjustment, just two sharp positions on a metre scale — which is why this remains the standard bench method and a fixture of JEE practical-style questions.
Why two positions exist is worth five lines. For fixed u + v = D, the lens equation 1/v + 1/u = 1/f can be satisfied by (u1, v1) and equally by the swapped (v1, u1) — object and image distances exchange roles, the principle of reversibility rendered on a bench. The two lens positions sit symmetric about the midpoint, separated by d = |v1 − u1|, and eliminating u, v yields f = (D² − d²)/(4D). When the positions merge (d = 0), the lens is midway, u = v = D/2, and D = 4f — the minimum separation for which any real image forms.
Where students slip
The separation formula's minus sign is the first trap: students write 1/F = 1/f1 + 1/f2 + d/(f1f2) and conclude separation strengthens a converging pair, opposite to the truth. Sign errors dominate mixed pairs — when f2 is negative, the product f1f2 is negative and the d-term actually adds power; the algebra carries the physics, so substitute with signs rather than reasoning in words. In displacement-method problems, the sanity check f < D/4 always (equality only when d = 0) catches mixed-up D and d. Stage-by-stage magnification loses marks when the intermediate image lands beyond the second lens (a virtual object for it) — draw the ray chain before writing the second lens equation, and remember m_total is the product regardless.
Frequently asked questions
What is the equivalent power of two thin lenses in contact?
P = P1 + P2, the dioptres adding algebraically, so a converging lens can be cancelled exactly by a suitable diverging one in contact.
How does separating two lenses change the combination?
The equivalent focal length follows 1/F = 1/f1 + 1/f2 − d/(f1f2), so a finite separation weakens a converging pair relative to contact — the basis of telephoto and zoom designs.
What is the displacement method for focal length?
With object and screen fixed a distance D apart, the lens focuses sharply at two positions separated by d, and f = (D² − d²)/(4D) — no focal-point location needed.
Why must object–screen distance exceed 4f?
Because u + v = D with the lens equation admits real solutions only when D ≥ 4f; at exactly 4f the two lens positions merge at the midpoint.
How is total magnification of a lens pair computed?
Stage by stage: the first lens's image is the second lens's object, and m_total = m1 × m2, with virtual intermediate objects handled by the sign convention rather than a new formula.