Two faces close together
A lens is a piece of glass, or of any transparent material, bounded by two refracting surfaces. At least one of them is curved. When the thickness of the lens is small compared with its radii and with the distances of object and image, it is a thin lens. Its two poles then nearly coincide, at one point called the optical centre. The line through the two centres of curvature is the principal axis of the lens.
A thin lens of glass in air does one of two things to a parallel beam. A converging lens brings the beam to a real focus. A diverging lens spreads the beam, so that it seems to come from a virtual focus. The shape tells you which. A glass lens in air that is thicker in the middle than at the edges converges. One that is thinner in the middle diverges.
Why does the thickness decide? Near its edge, each small part of a lens is like a thin prism. The chapter on Refraction showed that a prism turns a ray towards its base, the thicker side. In a lens thicker in the middle, every edge prism has its base towards the axis, so the rays bend towards the axis. In a lens thinner in the middle, the bases point outwards, and the rays bend away.
A thin lens with the same medium on both sides has two foci, \(F_1\) and \(F_2\), at equal distances from the optical centre. That distance is the focal length \(f\). On the Cartesian convention, a converging lens has \(f > 0\): its second focus lies ahead, along the light. A diverging lens has \(f < 0\).
A thin glass lens in air has two faces that curve the same way. The lens is thicker at its middle than at its edge. What does it do to a parallel beam?
Building the lens maker's formula
A thin lens is two refracting faces with nothing between them to speak of. So the method of the last lesson applies: the first face forms an image, and that image is the object for the second face. Because the lens is thin, the image of the first face is at the same distance from both poles. Fig 3.2 shows the two steps for an equiconvex glass lens of radii \(20\,\text{cm}\).
Put numbers to the two chips. First face, air to glass, \(R_1 = +20\,\text{cm}\): \(\dfrac{1.5}{v_1} = \dfrac{0.5}{20}\), so \(v_1 = +60\,\text{cm}\). Second face, glass to air, \(R_2 = -20\,\text{cm}\), with \(u_2 = +60\,\text{cm}\): \(\dfrac{1}{v} = \dfrac{1.5}{60} + \dfrac{-0.5}{-20} = 0.05\), so \(v = +20\,\text{cm}\). Each face does half the work.
Now do the same with symbols, for a lens of index \(\mu\) in a medium of index \(\mu_m\). The result is the lens maker's formula.
\[ \dfrac{\mu}{v_1} - \dfrac{\mu_m}{u} = \dfrac{\mu - \mu_m}{R_1} \]
Why. the first face takes light from the medium into the glass and forms an image at \(v_1\).
\[ \dfrac{\mu_m}{v} - \dfrac{\mu}{v_1} = \dfrac{\mu_m - \mu}{R_2} \]
Why. the second face takes light from the glass back into the medium. The lens is thin, so the first image is at \(v_1\) from this pole too.
\[ \mu_m\left(\dfrac{1}{v} - \dfrac{1}{u}\right) = (\mu - \mu_m)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \]
Why. add the two equations. The terms in \(v_1\) cancel, and \(\mu_m - \mu = -(\mu - \mu_m)\).
\[ \dfrac{1}{v} - \dfrac{1}{u} = \left(\dfrac{\mu}{\mu_m} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \]
Why. divide both sides by \(\mu_m\).
\[ \dfrac{1}{f} = \left(\dfrac{\mu}{\mu_m} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \]
Read it. a very distant object, \(u \to -\infty\), gives its image at the focus, \(v = f\). The right side depends only on the lens and the medium.
The signs do the work. An equiconvex lens of radii \(20\,\text{cm}\) has \(R_1 = +20\,\text{cm}\), with its centre ahead. Its second face has \(R_2 = -20\,\text{cm}\), with its centre behind. So \(\dfrac{1}{f} = 0.5\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = 0.05\) and \(f = +20\,\text{cm}\). For any equiconvex lens of glass \(1.5\), \(\dfrac{1}{f} = 0.5 \times \dfrac{2}{R} = \dfrac{1}{R}\), so \(f = R\). A flat face has \(R = \infty\), and its term \(1/R\) is zero.
A thin lens has the same focal length whichever face meets the light first. Turn the plano-convex lens round, and both radii change sign and swap places. The bracket \((1/R_1 - 1/R_2)\) keeps its value.
A glass meniscus lens (\(\mu = 1.5\)) has \(R_1 = +10\,\text{cm}\) and \(R_2 = +20\,\text{cm}\). Find its focal length and say whether it converges or diverges.
Both centres lie ahead of the lens, along the light, so both radii are positive, as given.
\(\dfrac{1}{f} = (1.5 - 1)\left(\dfrac{1}{10} - \dfrac{1}{20}\right)\).
\(\dfrac{1}{f} = 0.5 \times 0.05 = 0.025\), so \(f = +40\,\text{cm}\).
The lens converges, with \(f = +40\,\text{cm}\). Its more strongly curved face, of radius \(10\,\text{cm}\), is the convex one, so the lens is thicker in the middle.
An equiconvex lens with radii of \(30\,\text{cm}\) has a focal length of \(30\,\text{cm}\) in air. Find the refractive index of its glass.
A thin plano-convex glass lens (\(\mu = 1.5\)) has a focal length of \(30\,\text{cm}\). Find the radius of its curved face, in cm.
A lens under water
A lens bends light because its index differs from that of its surroundings. The formula shows this through the relative refractive index \(\mu/\mu_m\). Put the glass lens of \(20\,\text{cm}\) into water. Now \(\mu/\mu_m = 1.5/\tfrac43 = 1.125\), and the factor \(\mu/\mu_m - 1\) falls from \(0.5\) to \(0.125\). The radii are unchanged, so \(1/f\) falls to a quarter. The focal length becomes \(80\,\text{cm}\).
Dividing the two forms of the formula removes the radii. The focal length \(f'\) in a medium follows from the focal length \(f\) in air. \[ f' = f\,\dfrac{\mu - 1}{\mu/\mu_m - 1} \]
Watch the parallel beam as the index of the liquid rises. At \(1.25\) the focus is at \(50\,\text{cm}\), and in water at \(80\,\text{cm}\). At \(1.50\) the rays pass straight through: \(\mu/\mu_m = 1\), and \(1/f = 0\). No light is bent or reflected at the faces, so the lens cannot be seen in the liquid. Above \(1.50\) the lens diverges the beam. In a liquid of index \(1.6\) (a value chosen for this example), \(\mu/\mu_m = 0.9375\) and \(f = -160\,\text{cm}\).
An air bubble in water shows the same reversal. A bubble shaped like an equiconvex lens, with radii of \(10\,\text{cm}\), has \(\mu/\mu_m = 1/\tfrac43 = 0.75\). Then \(\dfrac{1}{f} = (0.75 - 1)\left(\dfrac{1}{10} + \dfrac{1}{10}\right) = -0.05\), so \(f = -20\,\text{cm}\). A "convex" bubble of air in water diverges light.
A glass lens (\(\mu = 1.5\)) has a focal length of \(20\,\text{cm}\) in air. Find its focal length in a liquid of refractive index \(1.25\), in cm.
An equiconvex glass lens (\(\mu = 1.5\)) of focal length \(20\,\text{cm}\) in air is placed in different liquids. Which statements are true?
Power, in dioptres
A lens with a short focal length bends light strongly. So opticians describe a lens by the power \(P = 1/f\), with \(f\) in metres. The unit of power is the dioptre, D, equal to \(1\,\text{m}^{-1}\). A converging lens has positive power, and a diverging lens negative power. The lens of \(20\,\text{cm}\) has \(P = 1/0.2 = +5\,\text{D}\). In water, with \(f = 0.8\,\text{m}\), it has only \(1.25\,\text{D}\).
The power of a lens is the sum of the powers of its two faces. For the equiconvex lens, each face has \((\mu - 1)/R = 0.5/0.2 = 2.5\,\text{D}\), and together they make \(5\,\text{D}\). This is why powers are convenient. Powers of lenses placed in contact also add, as a later lesson shows.
A spectacle prescription from the optician gives the power of each lens in dioptres, with its sign. A lens of \(-2.50\,\text{D}\) is a diverging lens with \(f = 1/(-2.5) = -0.40\,\text{m} = -40\,\text{cm}\), for an eye that cannot see far objects clearly. A reading lens of \(+2.00\,\text{D}\) converges, with \(f = +50\,\text{cm}\).
A spectacle prescription reads \(-4.00\,\text{D}\) for the right eye. What lens is this?
Until the 1870s, opticians labelled a lens by its focal length, often in inches. The inch itself differed from one country to another, so the same number could mean different lenses in different places. In 1872 the French eye doctor Ferdinand Monoyer proposed a unit based on the metre: the reciprocal of the focal length. He called it the dioptre. The unit has a second advantage: the powers of thin lenses held together simply add.
- Giving both radii of an equiconvex lens the same sign.Sign each radius from where its own centre lies. The two centres of an equiconvex lens are on opposite sides, so \(R_1 > 0\) and \(R_2 < 0\).
- Using \(\mu\) instead of \(\mu/\mu_m\) for a lens in a liquid.The lens bends light only as much as its index differs from its surroundings. In a liquid, use the relative index \(\mu/\mu_m\).
- Working out the power with \(f\) in centimetres.A dioptre is \(1\,\text{m}^{-1}\). Convert \(f\) to metres first: \(20\,\text{cm}\) gives \(5\,\text{D}\), not \(0.05\,\text{D}\).