Two sharp images from one lens
In the laboratory, the focal length of a convex lens is often found by the displacement method. A lit object and a screen are fixed a distance \(D\) apart. The lens slides along the bench between them, and the screen shows a sharp image only at certain places.
A lit object and a screen stand \(100\,\text{cm}\) apart. A convex lens of focal length \(21\,\text{cm}\) slides between them. At how many positions of the lens is the image on the screen sharp?
In the figure above the image lands on the screen twice. With the lens \(30\,\text{cm}\) from the object, \(v = +70\,\text{cm}\). With the lens \(70\,\text{cm}\) from the object, \(v = +30\,\text{cm}\). The two positions simply swap \(u\) and \(v\), because the lens formula treats the object and the image alike. Such a pair of positions, each the image of the other, are conjugate points. The two lens positions are \(d = 70 - 30 = 40\,\text{cm}\) apart.
\[ x + y = D, \qquad y - x = d \]
Why. let \(x = -u\) and \(y = v\) be the sizes of the two distances at the first position. Together they span \(D\). At the second position they swap, so the lens moves by \(y - x = d\).
\[ x = \dfrac{D - d}{2}, \qquad y = \dfrac{D + d}{2} \]
Why. add and subtract the two equations of step 1.
\[ \dfrac{1}{f} = \dfrac{1}{y} + \dfrac{1}{x} \]
Why. the lens formula \(1/v - 1/u = 1/f\) with \(v = y\) and \(u = -x\).
\[ \dfrac{1}{f} = \dfrac{2}{D + d} + \dfrac{2}{D - d} = \dfrac{4D}{D^2 - d^2} \]
Why. substitute step 2 and add the two fractions over the common denominator \(D^2 - d^2\).
\[ f = \dfrac{D^2 - d^2}{4D} \]
Read it. take the reciprocal. Only \(D\) and \(d\) are measured, never the distance to the optical centre.
For \(D = 100\,\text{cm}\) and \(d = 40\,\text{cm}\), \(f = \dfrac{10000 - 1600}{400} = 21\,\text{cm}\). The method has a practical advantage. A thick lens in a holder has no easily found optical centre. Here only the distance \(D\) and the movement \(d\) of the holder are measured, and neither needs the optical centre.
The two positions exist only if the screen is far enough away. From step 1, \(x + y = D\) and \(xy = fD\), so \(x\) and \(y\) are the roots of \(x^2 - Dx + fD = 0\). Real roots need \(D^2 \ge 4fD\), that is, \(D \ge 4f\). At \(D = 4f\) the two positions merge at \(u = -2f\), \(v = 2f\). For \(D < 4f\) no position gives a real image on the screen.
The two images give the size of the object too. At the first position \(|m_1| = y/x\), and at the second \(|m_2| = x/y\). So \(m_1 m_2 = 1\), and the image heights obey \(h_1 h_2 = h_o^2\). In the figure, \(m_1 = -2.33\) and \(m_2 = -0.43\), whose product is \(1\).
In a displacement experiment the two sharp images are \(4\,\text{cm}\) and \(1\,\text{cm}\) tall. How tall is the object, in cm?
The lens of the example has \(f = 21\,\text{cm}\). What is the least distance between object and screen for which it can form a real image on the screen, in cm?
When the edge focuses nearer
Every formula of this chapter used paraxial rays. A real lens also passes rays far from the axis, and for them the small-angle rules fail. Fig 6.1 traces a wide parallel beam through a thick convex lens, with Snell's law applied exactly at both faces. The rays near the axis meet at the paraxial focus \(F_p\). The marginal rays, through the edge of the lens, cross the axis nearer the lens, at \(F_m\).
This fault is spherical aberration. A sphere is simply not the ideal shape for focusing: its edge bends light more than the paraxial formula allows. So a point object gives a small blurred patch instead of a point image. The blur grows quickly with the width of the beam.
There are several cures. A stop, an opaque plate with a small opening, admits only the paraxial rays; the image sharpens, but it grows dimmer. A plano-convex lens with its curved face towards a parallel beam shares the bending between its two faces, and the fault is smaller. Two lenses can be combined so that their faults cancel. The small lenses of a phone camera are moulded with faces that are not spherical at all.
A stop with a small opening is placed in front of a convex lens. What happens to the image of a distant point?
Colours that focus apart
The chapter on Refraction showed that the refractive index of glass is larger for violet light than for red. The lens maker's formula then gives each colour its own focal length. Take an equiconvex lens of radii \(20\,\text{cm}\) with \(\mu = 1.55\) for violet and \(1.50\) for red. Then \(\dfrac{1}{f} = \dfrac{\mu - 1}{10}\), so \(f_V = \dfrac{10}{0.55} = 18.2\,\text{cm}\) and \(f_R = \dfrac{10}{0.5} = 20.0\,\text{cm}\). The colours of white light meet at points \(1.8\,\text{cm}\) apart.
This fault is chromatic aberration. On a screen at \(F_R\), the red light makes a sharp point, but the violet light has already crossed and spread out. The point is surrounded by a violet-blue fringe. The spread depends on the dispersive power \(\omega\) of the glass, the spread of the index divided by its mean deviation. \[ \omega = \dfrac{\mu_V - \mu_R}{\mu_Y - 1} \] With \(\mu_Y = 1.525\) for yellow, the glass above has \(\omega = 0.05/0.525 = 0.095\).
Differentiate the lens maker's formula to see how far apart the colours meet. With \(K = 1/R_1 - 1/R_2\), \(1/f = (\mu - 1)K\), so a change \(\Delta\mu\) changes \(1/f\) by \(K\Delta\mu = \omega/f\). The spread of the foci is then about \(\omega f\). For \(\omega = 0.02\) and \(f = 20\,\text{cm}\) it is \(0.4\,\text{cm}\). For the glass above, \(0.095 \times 19.05\,\text{cm} = 1.8\,\text{cm}\), as found.
The achromatic doublet
Place two thin lenses in contact. The power of the pair is \(1/F = 1/f_1 + 1/f_2\). Its spread between the colours is the sum of the two spreads, \(\omega_1/f_1 + \omega_2/f_2\). For an achromatic doublet, that sum must vanish.
Take a convex lens of crown glass, \(f_1 = +20\,\text{cm}\) and \(\omega_1 = 0.02\), and a lens of flint glass with \(\omega_2 = 0.04\). The condition gives \(\dfrac{0.02}{20} + \dfrac{0.04}{f_2} = 0\), so \(f_2 = -40\,\text{cm}\): a weaker diverging lens. The pair has \(\dfrac{1}{F} = \dfrac{1}{20} - \dfrac{1}{40} = \dfrac{1}{40}\), so \(F = +40\,\text{cm}\). It still converges, and now every colour meets at one focus.
The two glasses must differ. With the same glass, \(\omega_1 = \omega_2\), and the condition forces \(f_2 = -f_1\). Then \(1/F = 0\): the colours meet, but the pair focuses nothing. Flint glass spreads colour much more than crown glass for the same bending. So a weak flint lens can undo the colour spread of a stronger crown lens and leave most of its power.
A convex crown-glass lens of focal length \(25\,\text{cm}\) (\(\omega = 0.02\)) is to be made achromatic with a flint-glass lens (\(\omega = 0.04\)) in contact. Find the focal length of the flint lens, with its sign, in cm.
White light falls on a single convex glass lens. Which colour comes to a focus nearest the lens, and why?
Isaac Newton judged that a lens could never be freed from colour. He therefore built telescopes with mirrors instead, which reflect all colours alike. Around 1733 an English lawyer, Chester Moor Hall, quietly made lenses of crown and flint glass that showed little colour. In 1758 the London optician John Dollond patented the achromatic doublet and began to sell telescopes built with it. Nearly every good camera lens today is a descendant of that pair of glasses.
- Taking \(D\) as the distance from the lens to the screen.\(D\) is the fixed distance from the object to the screen, and \(d\) is how far the lens moves between the two sharp positions.
- Thinking the two sharp images have unrelated sizes.The two positions swap \(u\) and \(v\), so \(m_1 m_2 = 1\) and the object height is \(\sqrt{h_1 h_2}\).
- Building an achromat from two lenses of the same glass.The same glass gives \(f_2 = -f_1\) and zero power. The two lenses must have different dispersive powers.