Lens-Mirror Systems
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Direct answer
Mirrors image by 1/v + 1/u = 1/f with f = R/2, thin lenses by 1/v − 1/u = 1/f, both under the Cartesian sign convention: distances measured from the pole or optical centre, positive along the incident-light direction and negative against it. Systems are solved sequentially — each element's image becomes the next element's object, with careful sign transfer — and thin lenses in contact combine as 1/F = 1/f1 + 1/f2, powers adding directly.
What you must remember
- Mirror: 1/v + 1/u = 1/f, f = R/2; concave f negative, convex positive; magnification m = −v/u, negative meaning inverted.
- Thin lens: 1/v − 1/u = 1/f; convex f positive, concave negative; m = v/u; a lens's virtual image has positive v.
- Lens maker's formula 1/f = (n − 1)(1/R1 − 1/R2); power P = 1/f in metres (dioptres); lenses in contact: P = P1 + P2.
- A lens silvered on one face acts as a mirror of equivalent power 2 P(lens) + P(mirror) — the light crosses the lens twice.
- Two plane mirrors at angle theta give (360/theta − 1) images when 360/theta is an integer; plane images are virtual, erect and laterally inverted.
- Refraction: Snell's law n1 sin(theta1) = n2 sin(theta2); real depth = n times apparent depth; a slab of thickness t shifts the image by t(1 − 1/n).
- Total internal reflection, denser to rarer medium: sin C = n2/n1; beyond the critical angle light reflects totally — optical fibres, diamond brilliance and totally reflecting prisms.
Common confusion
Virtually every lost mark is a sign error: mixing conventions, forgetting u is negative for a real object, or forgetting that after reflection light travels backward so later distances are re-measured against the new direction. The second slip is mishandling a virtual object — when one element's image lies beyond the next element, it serves as a real-object-directioned virtual object there. Write every distance with its sign, from the correct origin, before substituting.
Exam-focused takeaway
JEE Main tests single-element numericals, lens combinations and power, apparent depth and critical angle — marks won purely on immaculate signs. JEE Advanced prefers layered systems: lens-plus-mirror setups (three passes or the silvered-lens equivalent), the displacement method f = (D^2 − d^2)/(4D) for fixed object–screen distance D and conjugate positions separated by d, a lens immersed in liquid (f scales with index contrast, and a converging lens can turn diverging), and u–v graphs. Solve element by element, transferring signs like currency.
Frequently asked questions
What sign convention does JEE use for mirrors and lenses?
The Cartesian one: distances measured from the pole or optical centre, positive along the incident light (usually rightward); u is negative for real objects, concave-mirror f negative, convex-lens f positive.
How is a lens–mirror combination solved?
Sequentially: the lens's image becomes the mirror's object (distances re-measured, light now reversed), then the reflected light crosses the lens again — or use equivalent power 2 P(lens) + P(mirror) when silvered.
What happens to a convex lens in water?
Its focal length grows, since what matters is the relative index between glass and surroundings; in a medium denser than glass the same shape becomes diverging.
What is the displacement method?
With object and screen fixed at separation D, sharp images form at two lens positions separated by d; then f = (D^2 − d^2)/(4D), from conjugate points.
When does total internal reflection occur?
Travelling denser to rarer at incidence beyond the critical angle C, with sin C = n2/n1 — no refracted ray then exists and reflection is total.