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Light has no mass, yet it pushes on whatever it hits. The push is tiny, but it is real, it has been measured, and spacecraft have used it. In problems it usually shows up as one line: find the force or pressure of a beam on a surface. This post shows where that force comes from, works through four problems with numbers, and lists the slips that cost marks.
Where does photon momentum come from?
The NCERT Class 12 chapter Dual Nature of Radiation and Matter sums up the photon picture: each photon has energy E = hν and momentum p = hν/c = h/λ. A brighter beam of the same wavelength does not have bigger photons. It has more photons crossing a given area each second.
For green light of wavelength 500 nm, using the CODATA value of the Planck constant, h = 6.626 × 10⁻³⁴ J s:
p = h/λ = 6.626 × 10⁻³⁴ / 5.00 × 10⁻⁷ = 1.33 × 10⁻²⁷ kg m/s
Is radiation pressure on the syllabus?
The Syllabus for JEE (Main) 2026 lists the "particle nature of light" in Unit 17 and "electromagnetic waves and their characteristics" in Unit 15. The current reprint of the NCERT chapter on electromagnetic waves says these waves carry energy but has no separate section on radiation pressure. So treat force-of-light problems as applications of photon momentum, and check the syllabus for your exam year.
How do you turn photon momentum into a force?
Force is momentum delivered per second. Take a beam of power P and wavelength λ.
Energy per photon: E = hc/λ. Photons per second: n = P/E.
If the surface absorbs each photon, it receives p = E/c, so F = n × p = (P/E)(E/c) = P/c.
If it reflects each photon straight back, the momentum changes from +p to −p, a change of 2p, so F = 2P/c.
The wavelength cancels. For a beam of intensity I (watts per square metre) on area A, the pressure is I/c on an absorber and 2I/c on a mirror.
Worked example 1. A laser pointer emits 1.0 mW at 650 nm (sample numbers), and a matt wall absorbs it. Take c = 3.0 × 10⁸ m/s.
E = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) / 6.5 × 10⁻⁷ = 3.06 × 10⁻¹⁹ J
n = 1.0 × 10⁻³ / 3.06 × 10⁻¹⁹ = 3.27 × 10¹⁵ photons per second
p = h/λ = 1.02 × 10⁻²⁷ kg m/s
F = n × p = 3.33 × 10⁻¹² N
Check with the shortcut: F = P/c = 1.0 × 10⁻³ / 3.0 × 10⁸ = 3.33 × 10⁻¹² N. Same answer.
How hard does sunlight push?
Kopp and Lean (2011) report that total solar irradiance during the 2008 solar minimum was 1360.8 ± 0.5 W per square metre, measured by the TIM instrument on NASA's SORCE satellite. Use I = 1361 W/m².
Worked example 2. Pressure of sunlight at Earth's distance, at normal incidence:
Black absorbing surface: I/c = 1361 / 3.0 × 10⁸ = 4.54 × 10⁻⁶ Pa
Perfect mirror: 2I/c = 9.07 × 10⁻⁶ Pa
E. F. Nichols and G. F. Hull were measuring this push in the lab more than a century ago. They published a preliminary report in Physical Review in 1901 and a second paper on the pressure due to radiation in 1903.
Worked example 3. Japan's space agency JAXA launched IKAROS, a craft "accelerated by radiation of the Sun", on 21 May 2010. JAXA describes its sail as a square about 20 m along the diagonal.
Side = 20/√2 = 14.1 m, so area A = 14.1² ≈ 200 m²
Ideal estimate, perfect mirror facing the Sun at Earth's distance: F = 2IA/c = 2 × 1361 × 200 / 3.0 × 10⁸ = 1.8 × 10⁻³ N
About 1.8 mN, roughly the weight of a 0.2 g object. A real sail is not a perfect mirror, so treat this as an upper estimate. With no fuel used, though, that small push keeps acting.
What changes when light hits at an angle?
Let the beam make angle θ with the normal to a mirror.
Each reflected photon's momentum changes only in the normal direction, by 2p cos θ.
A beam of cross-section A spreads over a larger surface area, A/cos θ, so fewer photons strike each square metre.
Both effects multiply: pressure = 2I cos²θ / c.
Worked example 4. Sunlight hits a mirror at θ = 60°. cos²60° = 0.25, so pressure = 2 × 1361 × 0.25 / 3.0 × 10⁸ = 2.27 × 10⁻⁶ Pa, a quarter of the normal-incidence value. For an absorber, the force along the normal gives I cos²θ / c, and there is also a component along the surface, because absorbed photons hand over their full momentum.
Common mistakes and how to avoid them
Dropping the 2 for reflection. Write the momentum before and after, with signs, every time.
Using cos θ once. One cos θ comes from the direction of the momentum change, the other from the spread-out beam.
Mixing up power and intensity. Intensity is in W/m². Multiply by area before you use F = P/c.
Thinking shorter wavelengths push harder at the same power. Each photon carries more momentum, but there are fewer of them. F = P/c does not depend on λ.
For more practice on photon energy and counting photons, see our post on photoelectric effect problems.
A practice routine for this topic
Derive F = P/c and F = 2P/c from photon counting once, on paper, without notes.
Solve one problem of each type: absorber, mirror, partial reflector and oblique mirror.
Check every answer's units, newtons for force and pascals for pressure, before you look at the key.
If the cos²θ step still feels like a trick, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Try it now: find the force of a 5 mW beam on a mirror and on black card, and explain why the answers differ by exactly a factor of 2.
