Tilt a phone towards a window and look at its camera lens. It shows a faint purple or green glint, not the white reflection of the window. New spectacles from the optician do the same. Look through a plain glass pane instead, and you see a clear, bright reflection of the room from both of its faces.
Glass that does not reflect
Bare glass reflects about \(4\,\%\) of the light at each face. A phone camera has several lenses, each with two faces. Every reflection loses light from the image and sends stray light across it, as a haze or a ghost of a bright lamp. So lenses are given an anti-reflection coating, a thin transparent layer on each face.
The coating has a refractive index between that of air and that of glass. A common one is magnesium fluoride, with \(\mu = 1.38\); this lesson uses that value and the glass value \(1.5\). Light meets two surfaces: air to coating, and coating to glass. Each is towards a denser medium, so both reflections gain a half wave, and the two half waves cancel.
With no half wave between them, the two reflected waves cancel when their path difference is half a wavelength. Seen straight on, that means \(2\mu t = \lambda/2\), so \(t = \lambda/4\mu\). The coating is a quarter of a wavelength thick, measured in the coating. For the middle of the visible range, \(552\,\text{nm}\), where the eye is most sensitive, \(t = 552/(4 \times 1.38) = 100\,\text{nm}\).
The two waves in Fig 3.1 are not quite equal, so they do not cancel completely. The front one has \(r = (1 - 1.38)/2.38 = -0.16\), the back one only \(-0.04\). What is left is about \(1.4\,\%\) of the light instead of \(4\,\%\). A perfect coating would need \(\mu = \sqrt{1.5} = 1.22\), which makes the two reflections equal; no hard, clear material has so low an index.
The coating is right for one wavelength only. At \(400\,\text{nm}\) and \(700\,\text{nm}\) its path difference is no longer half a wave, so it reflects a little more. It reflects about \(2.2\,\%\) of violet and \(1.7\,\%\) of red. Violet and red together look purple.
A coating of \(\mu = 1.25\) on glass of \(\mu = 1.5\) must cancel the reflection of \(600\,\text{nm}\) light at normal incidence. What is its least thickness, in nm?
A student says the anti-reflection coating on glass should be half a wavelength thick in the coating, like the least dark soap film. Why is this wrong?
Now return to the phone camera. Each lens face carries a coating about \(100\,\text{nm}\) thick. Its two reflections cancel near the middle of the spectrum. So most of the green and yellow light goes into the camera instead of being reflected. Violet and red are cancelled less well, and that weak remainder is the purple glint. A plain pane has no coating, so it reflects every colour.
The air film under a lens
Rest the curved face of a plano-convex lens on a flat glass plate. The lens touches the plate at one point O. Around it lies a thin film of air, zero at O and growing thicker outward. Its thickness at each place is fixed by the curve of the lens.
\[ r^2 + (R - t)^2 = R^2 \]
Why. Pythagoras' theorem in the right triangle of Fig 3.2, with C the centre of curvature and Q the point of the lens above the film.
\[ r^2 = 2Rt - t^2 \]
Why. expand \((R - t)^2 = R^2 - 2Rt + t^2\) and cancel \(R^2\).
\[ r^2 \approx 2Rt \]
Why. \(t\) is a few micrometres and \(R\) is about a metre, so \(t^2\) is negligible beside \(2Rt\).
\[ t = \dfrac{r^2}{2R} \]
Read it. divide by \(2R\).
Newton's rings
Let light of one wavelength fall straight down on the lens. Light reflects from the top of the air film (glass to air, no change) and from its bottom (air to glass, a half wave). So the air film is a film with one half wave. With \(\mu = 1\) and \(\cos r = 1\), it is dark where \(2t = n\lambda\).
Every point at a distance \(r\) from O has the same \(t\). So the dark places form circles round O, Newton's rings. Put \(t = r^2/2R\) into \(2t = n\lambda\): \(r^2/R = n\lambda\). The dark ring of order \(n\) has radius \(r_n = \sqrt{n\lambda R}\). Bright rings lie between, at \(r = \sqrt{(n - \tfrac12)\lambda R}\).
At the centre \(t = 0\), which is the \(n = 0\) dark condition. The centre is dark, however clean the contact, because only the half wave separates the two reflections there. In transmitted light the half wave is absent, and the centre is bright. The two patterns are complementary.
A lens with \(R = 1.6\,\text{m}\) rests on a plate and is lit normally with light of \(500\,\text{nm}\). Find the radius and the diameter of the fifth dark ring, and the thickness of the air film there.
\(r_5^2 = 5\lambda R = 5 \times 500 \times 10^{-9} \times 1.6 = 4 \times 10^{-6}\,\text{m}^2\), so \(r_5 = 2 \times 10^{-3}\,\text{m} = 2\,\text{mm}\).
The diameter is \(D_5 = 4\,\text{mm}\).
The film there is \(t = r^2/2R = 4 \times 10^{-6}/3.2 = 1.25 \times 10^{-6}\,\text{m} = 1.25\,\mu\text{m}\), five half waves of \(500\,\text{nm}\).
Look at the rings in the figure above. They are not evenly spaced; they crowd together outward. Each dark ring is half a wavelength of film thicker than the one inside it. But \(t\) grows as \(r^2\), so equal steps in \(t\) need smaller and smaller steps in \(r\). For \(R = 1.6\,\text{m}\) and \(500\,\text{nm}\) the radii are \(0.89\), \(1.26\), \(1.55\), \(1.79\) and \(2.00\,\text{mm}\): the gaps shrink from \(0.37\) to \(0.21\,\text{mm}\).
Now fill the gap with water, \(\mu = \tfrac43\). The dark condition becomes \(2\mu t = n\lambda\), so \(r_n^2 = n\lambda R/\mu\). Every radius shrinks by the factor \(1/\sqrt\mu = \sqrt{3/4} = 0.87\). The fifth dark ring moves in from \(2.00\) to \(1.73\,\text{mm}\). Water is denser than air but less dense than glass, so the half wave stays at the bottom surface and the centre stays dark.
A lens of radius of curvature \(1\,\text{m}\) rests on a flat plate, lit normally by light of \(600\,\text{nm}\). What is the radius of the fourth dark ring in reflected light, in mm?
Why is the centre of Newton's rings dark in reflected light, where the lens touches the plate?
Explain why Newton's rings get closer together as one moves outward from the centre.
Successive dark rings are places where the air film is thicker by the same step, half a wavelength. The thickness grows as the square of the distance from the centre, \(t = r^2/2R\). Near the centre the film thickens slowly, so a large step in \(r\) is needed; further out it thickens quickly, so a small step in \(r\) is enough. The radii go as \(\sqrt n\), and the gap between rings shrinks.
Measuring a wavelength with the rings
In a college physics lab, the lens and plate sit under a travelling microscope. A sodium lamp of \(589\,\text{nm}\) lights them through a glass plate held at \(45°\), which sends the light straight down. A sodium lamp gives nearly one wavelength, so twenty or thirty clear rings appear. The microscope reads the diameters of the rings.
The formula \(D_n^2 = 4n\lambda R\) could give \(\lambda\) from a single ring. In practice the contact is never perfect: a speck of dust or the pressure of the lens shifts every order by an unknown amount. So we take two rings and subtract: \[ D_m^2 - D_n^2 = 4(m - n)\lambda R. \] The unknown shift cancels, and only the number of rings between the two is needed.
With a lens of \(R = 1.6\,\text{m}\), the fifth and the twentieth dark rings have diameters \(4\,\text{mm}\) and \(8\,\text{mm}\). Find the wavelength of the light.
Use two rings so that any error at the contact cancels: \(D_{20}^2 - D_5^2 = 4(20 - 5)\lambda R\).
\(D_{20}^2 - D_5^2 = (64 - 16) \times 10^{-6} = 48 \times 10^{-6}\,\text{m}^2\).
\(\lambda = \dfrac{48 \times 10^{-6}}{4 \times 15 \times 1.6} = \dfrac{48 \times 10^{-6}}{96} = 5 \times 10^{-7}\,\text{m}\).
The wavelength is \(500\,\text{nm}\), blue-green light.
In a Newton's rings experiment with \(R = 1.5\,\text{m}\), the diameters of the fifth and the fifteenth dark rings are \(3.0\,\text{mm}\) and \(6.0\,\text{mm}\). Find the wavelength, in nm.
In the lab the same set-up is often used the other way round. The sodium wavelength, \(589\,\text{nm}\), is taken as known, and the rings measure \(R\). A student reads \(D_5 = 3.76\,\text{mm}\) and \(D_{15} = 6.51\,\text{mm}\). Then \(R = (6.51^2 - 3.76^2) \times 10^{-6}/(4 \times 10 \times 589 \times 10^{-9})\), which is \(1.20\,\text{m}\). Opticians' workshops test a lens surface in a similar way, against a master surface. Uneven rings show where the lens is out of shape by a fraction of a micrometre.
Robert Hooke had seen colours in thin flakes of mica in 1665. Isaac Newton went further. He pressed a lens of very large radius onto a flat glass and measured the coloured rings round the contact. He found that the squares of the diameters of the dark rings grew in equal steps, as the even numbers \(2, 4, 6, 8\). So the air at successive rings thickened in equal steps. He published this in his book Opticks in 1704. Newton thought light was a stream of particles and never took the rings as a sign of waves. Yet his careful numbers were the ones Thomas Young later used to measure the wavelength of light.
- Putting a half wave at both surfaces of the air film under a lens.The top reflection is glass to air, towards a rarer medium: no change. Only the bottom one, air to glass, gains \(\pi\).
- Using the radius of a ring when a diameter is given, or mixing mm with m.\(D_n^2 = 4n\lambda R\) and \(r_n^2 = n\lambda R\). Square the lengths in metres, or convert mm² by \(10^{-6}\).
- Expecting the rings to be evenly spaced.The radii go as \(\sqrt n\), so the rings crowd outward.
- Making a coating half a wave thick.A coating on glass has two half waves, which cancel. It cancels the reflection when it is a quarter wave thick, \(t = \lambda/4\mu_c\).