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A lens on a bench, a mirror some distance behind it, and an object in front: where is the final image? Students who handle one lens or one mirror easily often freeze here, because the light changes direction halfway. The method is a chain of single-element steps. This post sets out the rules and works two problems from start to finish.
Is this in the syllabus?
The JEE (Advanced) 2026 brochure lists "thin lenses; combinations of mirrors and thin lenses; magnification" under optics. The Syllabus for JEE (Main) 2026 lists the mirror formula, the thin lens formula and magnification, with "combination of thin lenses in contact" as the only named combination. The NEET (UG) 2026 syllabus has the same list as JEE Main. Separated systems use only those formulas. Check your year's documents.
The 2026-27 reprint of the NCERT Class 12 chapter Ray Optics and Optical Instruments uses the chain method in Example 9.8 for three separated lenses, noting that "though the image is real, it serves as a virtual object for the second lens". That sentence is the key to this topic.
What are the rules for each step?
Measure from the element you are working on. Distances are from that lens or mirror, never from the first one.
Positive is the direction the light is travelling at that step, following NCERT's Cartesian convention. After a reflection, the light goes the other way, so the positive direction flips.
Image becomes object. If the light reaches the next element before the image forms, the image is a virtual object for that element.
Count the passes. Lens, mirror and lens again is three steps.
Multiply magnifications. Total m = m₁ × m₂ × m₃.
Use 1/v − 1/u = 1/f for the lens with m = v/u, and 1/v + 1/u = 1/f for the mirror with m = −v/u.
Worked example 1: convex lens and plane mirror
An object stands 30 cm to the left of a convex lens of focal length 20 cm. A plane mirror is 50 cm to the right of the lens, facing it. Find the final image.
Step 1, the lens. Light travels left to right. u = −30 cm, f = +20 cm. 1/v = 1/20 + 1/(−30) = 1/60, so v = +60 cm, and m₁ = 60/(−30) = −2.
Step 2, the mirror. The lens image would be 60 cm to the right, but the mirror is only 50 cm away, so the light hits it while still converging towards a point 10 cm behind it: a virtual object. A plane mirror forms the image as far in front as the object is behind, so the image is 10 cm in front of the mirror, 40 cm to the right of the lens. m₂ = +1. The rays really meet there.
Step 3, the lens again. Light now travels right to left, so take right-to-left as positive. The object is 40 cm from the lens on the side the light comes from: u = −40 cm. 1/v = 1/20 − 1/40 = 1/40, so v = +40 cm, which is 40 cm to the left of the lens. m₃ = 40/(−40) = −1.
Result. The final image is real, 40 cm to the left of the lens, and m = (−2)(+1)(−1) = +2. It is the same way up as the object and twice the size.
Worked example 2: convex lens and concave mirror
An object stands 15 cm to the left of a convex lens of focal length 10 cm. A concave mirror of focal length 10 cm is 50 cm to the right of the lens, facing it.
Step 1, the lens. u = −15 cm, f = +10 cm. 1/v = 1/10 − 1/15 = 1/30, so v = +30 cm and m₁ = −2.
Step 2, the mirror. The image is 30 cm from the lens, so 20 cm in front of the mirror. Light reaches it before the mirror, so it is a real object: u = −20 cm, f = −10 cm. 1/v = −1/10 + 1/20 = −1/20, so v = −20 cm. The image is 20 cm in front of the mirror, at the same point as the object. m₂ = −(−20)/(−20) = −1.
Step 3, the lens again. The image is 30 cm from the lens, on the side the returning light comes from: u = −30 cm. 1/v = 1/10 − 1/30 = 1/15, so v = +15 cm, 15 cm to the left of the lens. m₃ = 15/(−30) = −0.5.
Result. The final image sits exactly where the object is, real, with m = (−2)(−1)(−0.5) = −1: inverted and the same size.
This is no coincidence. The intermediate image sits at the mirror's centre of curvature (2f = 20 cm), so the rays reflect back along their own paths.
Common mistakes and how to avoid them
Measuring from the first element. In example 1, the return-pass object is 40 cm from the lens, not 60.
Ignoring a virtual object. If the next element is closer than the image, the light never forms that image. It forms a virtual object behind that element.
Forgetting to flip the sign after reflection. Use the direction of travel at each step.
Stopping after the mirror. If the light comes back through the lens, there is a third step.
When the lens and mirror touch, the chain collapses into a single mirror, which our guide to silvered lens problems covers. For sign slips with a single element, see ray optics mistakes. If you get stuck on one step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Redo example 1 with the mirror 70 cm from the lens, so the lens image forms before the mirror.
Redo example 2 with the object 20 cm from the lens and find where the final image goes.
Solve NCERT Example 9.8 yourself, drawing one sketch per step, before reading its solution.
Try it now: in example 1, move the plane mirror to 30 cm from the lens and show that the reflected light converges to a point on the lens itself.
