Two lenses as one
Put two thin lenses in contact, on one axis. The first lens forms an image of the object. That image is the object for the second lens, at the same distance, since the lenses touch. Add the two lens formulas, and the combination behaves as a single thin lens.
\[ \dfrac{1}{v_1} - \dfrac{1}{u} = \dfrac{1}{f_1} \]
Why. the first lens forms an image of the object at \(v_1\).
\[ \dfrac{1}{v} - \dfrac{1}{v_1} = \dfrac{1}{f_2} \]
Why. that image is the object for the second lens. The lenses touch, so its distance is \(v_1\) from the second lens too.
\[ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{1}{F} \]
Read it. add the two lines. The terms in \(v_1\) cancel, leaving the lens formula of a single lens of focal length \(F\).
The combination has an equivalent focal length \(F\), and its power is the sum of the two powers, \(P = P_1 + P_2\). A convex lens of \(20\,\text{cm}\) and a concave lens of \(30\,\text{cm}\) in contact give \(\dfrac{1}{F} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{1}{60}\), so \(F = +60\,\text{cm}\). In dioptres, \(P = 5 - 3.33 = +1.67\,\text{D}\).
A convex and a concave lens of the same focal length cancel each other. For \(+20\,\text{cm}\) with \(-20\,\text{cm}\), \(P = 5 - 5 = 0\), and the pair acts as a flat plate of glass. Use the equivalent lens like any other. An object \(30\,\text{cm}\) from the pair of \(60\,\text{cm}\) gives \(\dfrac{1}{v} = \dfrac{1}{60} - \dfrac{1}{30} = -\dfrac{1}{60}\). So \(v = -60\,\text{cm}\): a virtual image on the object's side, with \(m = +2\).
Two thin lenses in contact have a combined focal length of \(12\,\text{cm}\). One of them has a focal length of \(20\,\text{cm}\). Find the focal length of the other, in cm.
At an eye test, the optician places lenses one behind another in a heavy trial frame on your nose. Each lens is marked with its power. A lens of \(+1.00\,\text{D}\) with one of \(+0.50\,\text{D}\) behind it makes \(+1.50\,\text{D}\), a lens of about \(67\,\text{cm}\). Because thin lenses in contact add their powers, the optician can try many powers with a small box of lenses. The prescription then records the total.
Two lenses with a gap
Now pull the two lenses apart by a distance \(d\). The light that leaves the first lens reaches the second at a different height, and the second lens bends it by a different amount. The equivalent focal length then depends on the gap. \[ \dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} - \dfrac{d}{f_1 f_2}, \qquad P = P_1 + P_2 - dP_1P_2 \]
Take a lens of \(20\,\text{cm}\) and one of \(30\,\text{cm}\), \(10\,\text{cm}\) apart. In dioptres, \(P_1 = 5\,\text{D}\), \(P_2 = 3.33\,\text{D}\) and \(d = 0.1\,\text{m}\). Then \(P = 5 + 3.33 - 0.1 \times 5 \times 3.33 = 6.67\,\text{D}\), and \(F = 1/6.67\,\text{m} = 15\,\text{cm}\). In contact, the same pair has \(P = 8.33\,\text{D}\) and \(F = 12\,\text{cm}\). The gap has weakened the pair.
The equivalent focal length \(F\) is not measured from either lens. It is measured from a plane called a principal plane, which lies elsewhere. Where the beam meets is found by the image-as-object rule. With the gap of \(10\,\text{cm}\), the first lens aims the beam at a point \(20\,\text{cm}\) away, \(10\,\text{cm}\) beyond the second lens. That point is a virtual object, \(u_2 = +10\,\text{cm}\). Then \(1/v = 1/30 + 1/10\), so the beam meets \(7.5\,\text{cm}\) beyond the second lens.
A converging lens of \(+20\,\text{cm}\) and a converging lens of \(+30\,\text{cm}\) stand \(5\,\text{cm}\) apart on one axis. Find their equivalent focal length, in cm.
A silvered lens is a mirror
Coat one face of a lens with silver, so that it reflects. Light now enters the lens, reflects from the silvered face, and leaves through the lens again. It crosses the lens twice and meets the mirror once. So a silvered lens behaves as a single mirror whose power is the sum of all three. \[ P = 2P_L + P_M \]
Here every power is counted positive when it converges the light. For the lens, \(P_L = 1/f_L\). For the silvered face, \(P_M = -1/f_M\), where \(f_M\) is its focal length as a mirror on the Cartesian convention. A concave mirror has \(f_M < 0\), so its \(P_M\) is positive. A flat face gives \(P_M = 0\). The whole system is a mirror of focal length \(f = -1/P\): a concave mirror when \(P > 0\), a convex one when \(P < 0\).
Take the plano-convex lens of \(f_L = 20\,\text{cm}\), silvered on its flat face. The flat mirror adds nothing, so \(P = \dfrac{2}{20} + 0 = 0.10\,\text{cm}^{-1}\). Then \(f = -\dfrac{1}{0.10} = -10\,\text{cm}\): the lens acts as a concave mirror of focal length \(10\,\text{cm}\). Silver the curved face instead, of radius \(10\,\text{cm}\). Seen from inside the glass it is a concave mirror, \(f_M = -5\,\text{cm}\), so \(P_M = 0.2\,\text{cm}^{-1}\). Then \(P = 0.1 + 0.2 = 0.3\,\text{cm}^{-1}\) and \(f = -3.3\,\text{cm}\).
Check the curved-face case by following the light. It enters through the flat face without bending, since the beam is parallel to the axis. The curved mirror, of radius \(10\,\text{cm}\), brings it towards a point \(5\,\text{cm}\) in front, inside the glass. Leaving through the flat face, that point appears at \(5/1.5 = 3.3\,\text{cm}\), the apparent depth. The formula and the tracing agree.
An object stands \(15\,\text{cm}\) in front of the plano-convex lens silvered on its flat face (\(f = -10\,\text{cm}\) as a mirror). Find the image.
Treat the system as a concave mirror, \(f = -10\,\text{cm}\), with \(u = -15\,\text{cm}\).
\(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\), so \(\dfrac{1}{v} = -\dfrac{1}{10} + \dfrac{1}{15}\).
\(\dfrac{1}{v} = -\dfrac{1}{30}\), so \(v = -30\,\text{cm}\). The magnification is \(m = -\dfrac{v}{u} = -\dfrac{-30}{-15} = -2\).
A real, inverted image forms \(30\,\text{cm}\) in front of the silvered lens, twice the size of the object.
An equiconvex glass lens of focal length \(20\,\text{cm}\), with both radii \(20\,\text{cm}\), is silvered on one face. Find the focal length of the equivalent mirror, as a size in cm.
A plano-concave lens of focal length \(-20\,\text{cm}\) is silvered on its flat face. What does it behave as?
Cutting a lens
The focal length of a lens depends only on the curvatures of its two faces and on its glass. So ask one question of every cut: does each piece keep both faces? Fig 5.3 shows an equiconvex lens of \(20\,\text{cm}\) cut two ways.
Cut along the axis, each half still has both curved faces, with the same radii. The lens maker's formula gives the same \(f = 20\,\text{cm}\). Only the aperture is halved. One half gathers half the light, so its image is complete but half as bright. Now pull the two halves slightly apart, across the axis. Each half has its own optical centre, so a point object gives two images, one from each half.
Cut across the middle, each piece keeps one curved face and gains a flat one. It is a plano-convex lens: \(\dfrac{1}{f} = 0.5 \times \dfrac{1}{20}\), so \(f = 40\,\text{cm}\). Put the two pieces back in contact, either way round. Two lenses of \(40\,\text{cm}\) in contact give \(\dfrac{1}{F} = \dfrac{1}{40} + \dfrac{1}{40}\), and \(F = 20\,\text{cm}\) again.
The lower half of a converging lens is covered with black paper. What happens to the image of a candle on a screen?
- Adding focal lengths instead of powers.For lenses in contact, add \(1/f\), or add the powers in dioptres. Then take the reciprocal.
- Counting the lens once in a silvered lens.The light passes through the lens on its way in and again on its way out, so \(P = 2P_L + P_M\).
- Thinking any cut halves the focal length, or doubles it.A cut along the axis leaves both faces, so \(f\) is unchanged. A cut across replaces a curved face by a flat one, so each piece has twice the focal length.