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Turn a mirror by a small angle and the spot of light it throws on a wall swings twice as far as you might expect. This doubling turns up in JEE questions, in an NCERT exercise on a galvanometer mirror, and in lab instruments that use a light beam as a pointer. This post proves the 2θ rule in three lines, works three problems with numbers and lists the slips that cost marks.
Is this in NCERT and the exam syllabus?
The NCERT Class 12 chapter Ray Optics and Optical Instruments (Reprint 2026-27) states the laws of reflection: the angle of reflection equals the angle of incidence, and the incident ray, reflected ray and normal lie in one plane. The text does not derive the rotating-mirror result, but Exercise 9.30 needs it: light falls normally on a mirror fixed to a galvanometer coil, the mirror turns by 3.5°, and you must find how far the spot moves on a screen 1.5 m away.
The Syllabus for JEE (Main) 2026 lists "Reflection of light, spherical mirrors, mirror formula" under Optics, and the NEET (UG) 2026 syllabus uses the same words. The JEE (Advanced) 2026 brochure lists "Reflection and refraction at plane and spherical surfaces". The 2θ rule is a direct use of the law of reflection, so it is fair game in all three. Check the syllabus for your exam year.
Why does the reflected ray turn through 2θ?
Keep the incident ray fixed and let it meet the mirror at angle of incidence i.
The reflected ray is deviated from the incident direction by δ = 180° − 2i.
Rotate the mirror by θ about an axis that lies in the mirror's surface and is perpendicular to the plane of incidence. The normal rotates by θ too, so the new angle of incidence is i + θ.
The new deviation is δ′ = 180° − 2(i + θ). The change is δ − δ′ = 2θ.
So the reflected ray turns by 2θ, in the same sense as the mirror, whatever the starting angle. Check with numbers: at i = 30° the deviation is 120°. Turn the mirror 10° so that i = 40°, and the deviation is 100°. The reflected ray has moved 20°.
Two results follow at once. If the mirror turns at angular speed ω, the reflected beam sweeps round at 2ω. And the image of a fixed object also turns through 2θ about the axis of rotation, because the image always lies on the reflected rays traced backwards.
Worked example 1: an optical lever
A beam falls normally on a small mirror and returns along its own path to a scale 2.0 m away. The mirror turns by 2.0° (sample numbers). How far does the spot move?
The reflected ray turns by 2 × 2.0° = 4.0°.
Spot displacement = 2.0 × tan 4.0° = 0.140 m, or 14.0 cm.
Small-angle check: 2.0 m × (4.0° in radians, 0.0698) = 0.1396 m. The two agree to within 0.3 mm.
That is why this set-up is called an optical lever: a 2° turn becomes a 14 cm movement you can read with a ruler.
Worked example 2: working backwards
A spot on a scale 1.0 m from a mirror moves by 6.0 cm. By what angle did the mirror turn?
The reflected ray turned by tan⁻¹(0.060/1.0) = 3.43°.
The mirror turned by half of that: 1.72°.
The most common wrong answer is 3.43°, which forgets to halve.
Worked example 3: a spinning mirror and a moving spot
A plane mirror spins at 30 revolutions per minute about a vertical axis, with a fixed laser beam striking it. A wall is 5.0 m away (sample numbers).
Mirror angular speed ω = 30 × 2π/60 = 3.14 rad/s, so the beam sweeps at 2ω = 6.28 rad/s.
Where the beam meets the wall at right angles, the spot moves at 2ω × 5.0 = 31.4 m/s.
When the beam makes angle φ = 30° with the perpendicular to the wall, the spot sits at 5.0 tan φ along the wall, so its speed is 2ω × 5.0/cos²φ = 31.4/0.75 = 41.9 m/s.
The spot speeds up as it moves away from the foot of the perpendicular, even though the mirror spins steadily.
What mistakes cost marks?
Writing θ instead of 2θ. The normal turns by θ, the angles of incidence and reflection both change by θ, so the ray turns by 2θ.
Halving the wrong quantity. Given a spot movement, find the beam angle first, then halve it to get the mirror angle.
Mixing up rotation and translation. Moving a plane mirror a distance d along its normal moves the image by 2d. That is a different doubling; our post on image velocity in plane and spherical mirrors works through it.
Choosing the wrong axis. The rule holds for rotation about an axis in the mirror's plane, perpendicular to the plane of incidence. Spinning a flat mirror about its own normal leaves the reflected ray unchanged.
Using the small-angle form for large angles. Use tan for the spot position once the beam angle passes about 10°.
Two mirrors set at an angle give a related problem, the number of images. See images formed by two inclined plane mirrors.
A practice routine
Prove δ − δ′ = 2θ on paper, then repeat it for a rotation in the opposite sense.
Solve NCERT Exercise 9.30. (Answer: the beam turns 7.0°, so the spot moves 1.5 × tan 7.0° ≈ 18.4 cm.)
Redo Example 3 for φ = 45° and explain why the spot speed is exactly double its value at the foot of the perpendicular.
Write one optical-lever problem of your own, swap with a friend and check each other's halving step.
If the turning beam is hard to picture from a diagram, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, shine a phone torch on a small mirror, tilt the mirror slightly and measure how far the spot on the wall moves.
