Lesson
The theory — Thin-Film Interference
Thin-film interference: light reflected from the two surfaces of a layer comes back as two waves, the second delayed by δ = 4πn1d/λ for a layer of index n1 and thickness d at normal incidence. Each surface reflects an amplitude r = (na − nb)/(na + nb), negative where light goes into a denser material. Adding the waves, with every bounce inside the film, gives the reflectance R = (r01² + r1s² + 2r01r1s cos δ)/(1 + r01²r1s² + 2r01r1s cos δ).
What each symbol means
d- the film's thickness, in nm
n_1- the film's refractive index at the wavelength in question; water is 1.3347 at 550 nm
δ- how far the light from the bottom surface falls behind, in radians: 2π is one whole wavelength
Y- the luminous reflectance in white light: the reflected spectrum weighted by the eye's sensitivity, 100% for a perfect white reflector
- Assumes
- Light arriving head-on, a film of even thickness over the patch being looked at, and no absorption. A soap film is treated as water; the soap layers on each face are not modelled. White light is CIE illuminant D65.
Problem solved in full
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How many glass surfaces halve the light? 3 steps
A camera lens has many glass surfaces in a row. How many uncoated N-BK7 surfaces does it take for half the light to be lost to reflection, and how many coated ones? Count only the light that passes every surface without being reflected. Press Coated lens: the page gives the bare surface's 550 nm reflectance and the coated one's.
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Each surface passes 1 − R of what reaches it, so k surfaces pass (1 − R)k. Set that to one half:
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For bare glass, with R = 4.239%:
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For the coated surface, with R = 1.247%:
Answer
Seventeen bare surfaces, or 56 coated ones. The working gives k = 16.00 and 55.24, and surfaces come whole: sixteen bare ones still pass 50.01%, so it takes nine uncoated lens elements in air, not eight, to lose half of the 550 nm light. Ten surfaces pass 64.8% bare against 88.2% coated. The light that is lost does not vanish: it bounces between the surfaces and some of it reaches the picture as haze. Zeiss's 1935 patent gives that fogging of the negative as its second reason for the coating.
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Learning path
Wave optics
References (3)
- Newton's own account of a soap bubble under a glass cover: the coloured rings descending as the water subsides, and the black spot that grows at the top, from which he first thought no light was reflected (Obs. 17); the order of the colours as the bubble thins, the early ones "very dilute and dirty" (Obs. 18); and the opening of the book, that thin plates show colours "according to their various thinness, altho' at a greater thickness they appear very clear and colourless": I. Newton, Opticks: or, A Treatise of the Reflections, Refractions, Inflections and Colours of Light, 4th edition. William Innys, London, 1730. Book II, Part I, opening paragraph and Observations 17–18.
- Carl Zeiss's patent on raising the transmission of optical parts with a layer of lower index evaporated onto the glass in a vacuum, applied for on 1 November 1935; it gives Fresnel's reflectance, says that reflected and re-reflected light fogs a photographic negative, names calcium fluoride as the layer, and describes Taylor's British patent of 1904 as its forerunner, prompted by old lenses whose weathered patches looked more transparent: Carl Zeiss, Verfahren zur Erhöhung der Lichtdurchlässigkeit optischer Teile durch Erniedrigung des Brechungsexponenten an den Grenzflächen dieser optischen Teile, German patent DE 685767, filed 1 November 1935, published 23 December 1939.
- The CIE 1931 2° colour-matching functions and the spectrum of standard illuminant D65, the two tables every colour on this page is computed from, taken at 5 nm from 380 to 780 nm: CIE, Colour-matching functions of CIE 1931 standard colorimetric observer (dataset CIE_xyz_1931_2deg.csv), 2019, doi:10.25039/CIE.DS.xvudnb9b; and CIE standard illuminant D65 (dataset CIE_std_illum_D65.csv), doi:10.25039/CIE.DS.hjfjmt59. International Commission on Illumination, Vienna.