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Spherical Surfaces and Lenses · Lesson 4 of 6

The Lens Formula and Magnification

15 min read4 ideas5 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 4
The lens formula

The lens formula

Step 4 of the lens maker's derivation already held the answer for any object. Its right side is \(1/f\), so for every object distance \(u\) the image distance \(v\) obeys the lens formula. \[ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \] Both distances are measured from the optical centre, with the Cartesian signs. A converging lens has \(f > 0\), and a diverging lens \(f < 0\).

RememberThin lens formula
\[ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \]
Holds whena thin lens with the same medium on both sides, paraxial rays, Cartesian signs; a real object has \(u < 0\).
Symbols\(u\) object distance; \(v\) image distance; \(f\) focal length, all from the optical centre
Units\(u, v, f\) in the same unit, usually cm

Compare it with the mirror formula of the chapter on Reflection, \(1/v + 1/u = 1/f\). The sign between the two terms differs because a lens lets the light go on, while a mirror sends it back. A real image of a lens lies across the lens, with \(v > 0\). A virtual image lies on the object's side, with \(v < 0\).

Take a converging lens with \(f = +20\,\text{cm}\) and an object \(30\,\text{cm}\) in front of it. Then \(\dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{-30} = \dfrac{1}{60}\), so \(v = +60\,\text{cm}\). The image is real, \(60\,\text{cm}\) beyond the lens. Move the object to \(10\,\text{cm}\): \(\dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{10} = -\dfrac{1}{20}\), so \(v = -20\,\text{cm}\). The image is now virtual, on the object's side.

ExampleThe camera of a phone

The lens of a phone camera has a focal length of \(4.0\,\text{mm}\). A friend \(1.7\,\text{m}\) tall stands \(2.0\,\text{m}\) away. How far behind the lens must the sensor be, and how tall is the image?

Reveal one step at a time and finish the rest yourself.
Checkpoint · numericUnanswered

An object stands \(25\,\text{cm}\) in front of a converging lens of focal length \(20\,\text{cm}\). Find the image distance \(v\), with its sign, in cm.

cm± 2 %
Magnification of a lens
Maths you need

Magnification

A ray through the optical centre goes straight on, because the middle of a thin lens is like a thin parallel slab. The object, the image and the optical centre therefore lie on one straight line. Similar triangles on either side of the centre give the linear magnification. \[ m = \dfrac{h_i}{h_o} = \dfrac{v}{u} \]

RememberMagnification of a thin lens
\[ m = \dfrac{h_i}{h_o} = \dfrac{v}{u} \]
Holds whena thin lens, Cartesian signs; \(m < 0\) means a real, inverted image and \(m > 0\) a virtual, erect image.
Symbols\(h_o\) height of the object; \(h_i\) height of the image; \(m\) linear magnification
Units\(m\) is a pure number

For the object at \(30\,\text{cm}\), \(m = \dfrac{60}{-30} = -2\): the image is inverted and twice as tall. For the object at \(10\,\text{cm}\), \(m = \dfrac{-20}{-10} = +2\): the image is erect and twice as tall. There is no minus sign in front, unlike the mirror's \(m = -v/u\). For a single lens and a real object, a real image always has \(m < 0\) and a virtual image \(m > 0\).

Checkpoint · numericUnanswered

A stamp collector holds a magnifying glass of focal length \(5\,\text{cm}\) at \(4\,\text{cm}\) from a stamp. Find the magnification.

± 2 %
CheckpointUnanswered

An object stands \(30\,\text{cm}\) in front of a diverging lens of focal length \(20\,\text{cm}\). What is the image?

Ray diagrams for lenses

Three rays that find the image

Three rays from the top of an object are easy to draw. A ray parallel to the axis leaves through the second focus \(F_2\), or, for a diverging lens, seems to come from its first focus. A ray through the optical centre goes straight on. A ray through the first focus \(F_1\), or aimed at it, leaves parallel to the axis. Any two of them fix the top of the image.

Lens ray diagramA thin lens stands on the principal axis with its two foci F at 20 cm on each side and the points 2F at 40 cm. An upright object arrow, 8 cm tall, stands to the left at a distance set by a slider from 5 to 70 cm. Chips choose a convex lens, f = +20 cm, or a concave lens, f = −20 cm. Two rays leave the head of the object: one parallel to the axis, which the lens bends through the focus or as if from the focus, and one through the optical centre, which goes straight on. Where the rays meet, or where their dashed backward extensions meet, stands the image arrow, solid when it is real and dashed when it is virtual. A readout gives u, v and m. F F 2F 2F u = −30 cm, v = +60 cm, m = −2
SimA thin lens with its foci F at \(20\,\text{cm}\) and the points 2F at \(40\,\text{cm}\). Choose a convex or a concave lens with the chips, and slide the object from \(5\) to \(70\,\text{cm}\). A real image is solid, and a virtual image is dashed, with dashed backward extensions of the rays. The readout gives \(u\), \(v\) and \(m\).

Slide the object towards a convex lens and read the image. At \(60\,\text{cm}\), beyond 2F, the image is at \(+30\,\text{cm}\), with \(m = -0.5\). At \(40\,\text{cm}\), on 2F, it is at \(+40\,\text{cm}\), with \(m = -1\). At \(25\,\text{cm}\) it is far away, at \(+100\,\text{cm}\), with \(m = -4\). At F the rays leave parallel and the image is at infinity. Inside F the rays spread, and a large erect virtual image appears on the object's side.

Property
Object, convex lens
Image
Far away
\(u \to -\infty\)
at \(F_2\); real, inverted, a point
Beyond \(2F_1\)
\(u = -60\,\text{cm}\)
between \(F_2\) and \(2F_2\), \(v = +30\,\text{cm}\); real, inverted, smaller, \(m = -0.5\)
At \(2F_1\)
\(u = -40\,\text{cm}\)
at \(2F_2\), \(v = +40\,\text{cm}\); real, inverted, the same size
Between \(F_1\) and \(2F_1\)
\(u = -30\,\text{cm}\)
beyond \(2F_2\), \(v = +60\,\text{cm}\); real, inverted, larger, \(m = -2\)
At \(F_1\)
\(u = -20\,\text{cm}\)
at infinity
Inside \(F_1\)
\(u = -10\,\text{cm}\)
on the object's side, \(v = -20\,\text{cm}\); virtual, erect, larger, \(m = +2\)

Now choose the concave lens. Wherever the object stands, the rays leave spreading, and the image is virtual, erect and smaller. It always lies between the lens and F on the object's side. At \(30\,\text{cm}\) the readout gives \(v = -12\,\text{cm}\) and \(m = +0.40\). At \(60\,\text{cm}\) it gives \(v = -15\,\text{cm}\) and \(m = +0.25\).

CheckpointUnanswered

Where must an object stand, in front of a convex lens of focal length \(f\), to give a real image of its own size?

In the world
Why slides go in upside down

A projector in a classroom holds its slide just beyond the focus of a converging lens. Let the lens have \(f = 10\,\text{cm}\) and the slide stand \(10.2\,\text{cm}\) from it. Then \(1/v = 1/10 - 1/10.2\), so \(v = 510\,\text{cm}\), and \(m = 510/(-10.2) = -50\). The screen, \(5.1\,\text{m}\) away, shows the slide fifty times larger and inverted. So the slide is loaded upside down, and the picture on the screen stands upright.

Graphs of v against u for a lens
Maths you need

The graph of v against u

Solve the lens formula for \(v\): \(v = \dfrac{uf}{u + f}\). For a convex lens of \(f = +20\,\text{cm}\), Fig 4.1 plots \(v\) against \(u\) for real objects, \(u < 0\). The curve has two branches, a real branch and a virtual branch, separated by the line \(u = -f\).

Image distance against object distance for a convex lensA graph of the image distance v against the object distance u for a convex lens of focal length 20 cm, both in centimetres, u negative along the horizontal axis. The curve has two branches. For u beyond −20 cm, v is positive: a real image, very far for an object just beyond the focus and falling towards 20 cm for a distant object. For u between −20 cm and zero, v is negative: a virtual image. Dashed lines at u = −20 cm and v = +20 cm are the asymptotes. Marked points: (−30, 60), (−40, 40), where the image is the size of the object, and (−15, −60).−20−40−60−8020406080100−20−40−60−80Ou (cm)v (cm) (−30, 60) (−15, −60) real images virtual images v = f
Fig 4.1\(v\) against \(u\) for a convex lens of focal length \(20\,\text{cm}\). The dashed lines at \(u = -20\,\text{cm}\) and \(v = +20\,\text{cm}\) are the asymptotes. The dot at \((-40, 40)\) is the object at 2F, whose image is the same size; the other dots are \((-30, 60)\) and \((-15, -60)\).

Read the two branches. For \(u\) beyond \(-20\,\text{cm}\), \(v\) is positive: the images are real. A distant object puts the image near \(v = +20\,\text{cm}\), and an object just beyond F throws it very far. For \(u\) between \(-20\,\text{cm}\) and zero, \(v\) is negative: the images are virtual. At the point \((-40, 40)\), \(u = -2f\) and \(v = 2f\). There the object and its image are nearest to each other, \(4f = 80\,\text{cm}\) apart.

Two more graphs are useful. Plot \(1/v\) against \(1/u\): the formula \(1/v = 1/u + 1/f\) is a straight line of slope \(1\), crossing the \(1/v\) axis at \(1/f\). Laboratories use it to find \(f\) from a set of readings. The magnification also follows from \(u\) alone: \(m = \dfrac{v}{u} = \dfrac{f}{u + f}\). At \(u = -30\,\text{cm}\) it gives \(m = \dfrac{20}{-10} = -2\), as the ray diagram showed.

CheckpointUnanswered

A student measures pairs of \(u\) and \(v\) for a convex lens. Which plot of the readings gives a straight line?

Exam traps
  1. Using the mirror formula \(1/v + 1/u = 1/f\) for a lens.
    For a lens the light goes on, and the formula is \(1/v - 1/u = 1/f\).
  2. Writing \(u\) as a positive number for a real object.
    A real object stands against the incident light, so \(u < 0\). Put the sign in before solving.
  3. Writing \(m = -v/u\) for a lens.
    The lens formula for magnification has no minus sign: \(m = v/u\). A real image then comes out with \(m < 0\).