Image Formation in the Camera – Optical Instruments (Senior 4 Physics) | Ganzaa

Optical Instruments

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Senior 4 · Physics

Optical Instruments

4 lessons

Lesson 1 of 4

Image Formation in the Camera

25 min

A box that remembers light

Every phone now carries a camera, and photographers at a wedding in Kigali or a football match at Amahoro Stadium use the same physics as the first box cameras of the nineteenth century: a converging lens forms a real image on a light-sensitive surface. What the photographer actually controls is the lens position, the size of the opening that lets light in, and how long it stays open. In this lesson you use the lens formula from Unit 1 to explain all three.

The parts of a camera and the image it forms

A camera (Figure 1.1) is a light-tight box with a converging lens at the front and a light-sensitive surface at the back: photographic film in older cameras, and an electronic image sensor (a grid of millions of tiny light detectors, each giving one pixel) in digital ones.

Figure 1.1: The main parts of a camera. The image on the sensor is real, inverted and diminished.

Part	What it does
converging lens	forms a real, inverted, diminished image of objects beyond 2F
focusing mechanism	moves the lens towards or away from the sensor so that the image falls exactly on it
diaphragm (aperture stop)	a ring of overlapping blades that sets the diameter $D$ of the opening, controlling how much light enters per second
shutter	opens for a set time, the exposure time $t$ , then closes
sensor or film	records the image

Most objects you photograph are many focal lengths away, so they lie beyond 2F, and the lens formula puts the image just beyond F: real, inverted and much smaller than the object. That is why a camera lens has a short focal length (a few millimetres in a phone, about 50 mm in a classic film camera) and why the box can be small.

Definition

The f-number $N$ of a lens is its focal length divided by the diameter of its aperture: $N = \frac{f}{D}$ . It is written f/ $N$ , for example f/2.8. The exposure of the sensor is the light energy it receives per unit area; it is proportional to $\frac{t}{N^2}$ , where $t$ is the exposure time. Example: a 50 mm lens set to f/2.8 has an aperture $D = \frac{50}{2.8} = 18\,\mathrm{mm}$ across.

Focusing: moving the lens

The sensor is fixed, so for each object distance $u$ the lens must sit at the right distance $v$ from the sensor. From $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ : for a very distant object $v = f$ ; as the object comes closer, $v$ increases, so the lens moves away from the sensor. The nearer the object, the more the lens must move. (Your eye solves the same problem differently: the distance from lens to retina is fixed, so it changes the focal length of its lens instead. See lesson 4.)

Worked example 1: focusing a 50 mm lens

A camera has a lens of focal length 50.0 mm. How far must the lens move from its position for distant objects, to focus on (a) an object 1.00 m away, (b) a flower 0.300 m away?

\text{(a)}\;\; v = \frac{1}{\frac{1}{50.0} - \frac{1}{1000}} = 52.6\,\mathrm{mm}, \quad\text{so the lens moves } 2.6\,\mathrm{mm}

\text{(b)}\;\; v = \frac{1}{\frac{1}{50.0} - \frac{1}{300}} = 60.0\,\mathrm{mm}, \quad\text{so the lens moves } 10.0\,\mathrm{mm}

Focusing on everything from 1 m to infinity needs only a few millimetres of travel; close-up ("macro") photography needs much more.

Worked example 2: how big is the image?

The same camera photographs a 1.70 m tall person standing 5.00 m away. Will the whole person fit on a sensor 24 mm high?

v = \frac{1}{\frac{1}{50.0} - \frac{1}{5000}} = 50.5\,\mathrm{mm}, \qquad h_i = h_o\,\frac{v}{u} = 1700\,\mathrm{mm} \times \frac{50.5}{5000} = 17.2\,\mathrm{mm}

Yes: the image is 17.2 mm tall, inverted, and fits on the 24 mm sensor with room to spare.

Aperture, f-number and exposure

The light energy reaching the sensor each second is proportional to the area of the aperture, $\frac{\pi D^2}{4}$ . The image area, for a distant scene, is proportional to $f^{2}$ (every dimension of the image scales with $v \approx f$ ). So the light per unit area of the image, which is what the sensor responds to, is proportional to $\frac{D^2}{f^2} = \frac{1}{N^2}$ . Multiply by the time the shutter is open:

\text{exposure} \propto \frac{t}{N^2}

The f-numbers on a lens form the series 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22. Each is about $\sqrt{2}$ times the one before, so each step ("one stop") halves the aperture area and the light admitted. A large f-number means a small aperture. Because the f-number already includes the focal length, f/8 gives the same exposure on any lens, which is why photographers use it rather than the diameter.

Figure 1.2: Depth of field. A narrow aperture makes thinner cones of light, so the blur from points that are not exactly in focus is smaller.

Depth of field. Only objects at one distance are focused exactly on the sensor. Light from a nearer or farther point meets in front of or behind it and makes a small disc, the circle of confusion, on the sensor. If that disc is smaller than one pixel (or than the eye can see in the final print), the point still looks sharp. With a narrow aperture the cones of light are thin and the discs small, so a large range of distances looks sharp: a large depth of field. A wide aperture gives a shallow depth of field, which portrait photographers use to blur the background. The cost of a narrow aperture is less light, so a longer exposure time, and moving subjects then blur.

Worked example 3: trading aperture for time

A correct exposure is f/8 at $\frac{1}{125}\,\mathrm{s}$ . A footballer is running, so the photographer wants $\frac{1}{500}\,\mathrm{s}$ to freeze the motion. What f-number gives the same exposure?

\frac{t_1}{N_1^2} = \frac{t_2}{N_2^2} \;\Rightarrow\; N_2 = N_1\sqrt{\frac{t_2}{t_1}} = 8 \times \sqrt{\frac{0.0020}{0.0080}} = 8 \times \sqrt{\frac{1}{4}} = 4

Open up to f/4 (two stops, four times the area). The depth of field becomes shallower, so the focus must be set carefully on the player.

Experiment 1.1: a model camera

Apparatus: a converging lens ( $f$ about 10 cm) in a holder, a sheet of tracing paper as a screen, a metre rule, a lit object such as a window or a lamp with a cross-wire, in a darkened room.

Point the lens at a window on the far side of the room or, better, at a distant scene outside. Move the tracing paper until the image is sharp. Measure $v$ : this is close to $f$ .
Place the lit object at 50 cm, 40 cm, 30 cm and 20 cm from the lens; each time move the screen to focus and measure $v$ . Notice that the screen must move away from the lens as the object comes closer.
Cover the lens with a card that has a small hole (about 5 mm) and look again: the image is dimmer, but objects at different distances are sharp together.

Steps 1 and 2 test the lens formula; step 3 shows that a smaller aperture gives a dimmer image with a greater depth of field.

Warning

Safety. Never point the lens at the Sun or look at the Sun through it: the focused sunlight can set paper on fire and permanently damage your eyes. Handle a hot lamp housing with care.

Warning

Common mistakes. (1) Thinking a larger f-number means a larger aperture: $N = \frac{f}{D}$ , so f/16 is a small opening. (2) Saying the lens moves towards the sensor to focus on near objects: it moves away, because $v$ increases as $u$ decreases. (3) Forgetting that exposure depends on $N^{2}$ : going from f/4 to f/8 cuts the light to a quarter, not a half.

Tip

One stop = a factor of 2 in light. Halving the time or moving one f-number up the series (×1.4) each lose one stop; to keep the same exposure, every stop lost on one must be regained on the other.

Aperture, explained: the mysterious f-number (Emmett Budd). Watch why the f-number is the focal length divided by the diameter of the opening, why the series 1.4, 2, 2.8, 4 goes up by about √2, and how aperture changes the background blur.

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Practice

List the parts of a camera and state the job of each.
A phone camera has a lens of focal length 4.0 mm and an aperture 2.0 mm across. Find its f-number.
A camera lens of focal length 35 mm is focused on an object 2.0 m away. How far is the lens from the sensor?
A tree 12 m tall is photographed from 60 m with a 50 mm lens. Find the height of the image on the sensor.
A correct exposure is f/5.6 at $\frac{1}{250}\,\mathrm{s}$ . Find the exposure time needed at f/11.
Explain why landscape photographers often use f/16 while portrait photographers often use f/2.
A student says: "A camera and an eye both focus by moving the lens." Evaluate this statement.

Show answersHide answers

Lens: forms a real image; focusing ring: moves the lens to put the image on the sensor; diaphragm: sets the aperture and so the light per second; shutter: sets the exposure time; sensor or film: records the image; light-tight box: keeps out stray light.
$N = \frac{4.0}{2.0} = 2.0$ , so f/2.
$v = \frac{1}{\frac{1}{35} - \frac{1}{2000}} = 35.6\,\mathrm{mm}$ .
$v = \frac{1}{\frac{1}{50} - \frac{1}{60\,000}} = 50.04\,\mathrm{mm}$ ; $h_i = 12\,000 \times \frac{50.04}{60\,000} = 10\,\mathrm{mm}$ .
$t_2 = t_1 \left(\frac{N_2}{N_1}\right)^2 = \frac{1}{250} \times \left(\frac{11}{5.6}\right)^2 = 0.0154\,\mathrm{s} \approx \frac{1}{60}\,\mathrm{s}$ (two stops, four times as long).
f/16 is a small aperture, so the depth of field is large and both the near foreground and the distant hills are sharp; the scene is still, so the long exposure does not matter. f/2 gives a shallow depth of field, so the face is sharp and the background is blurred, making the person stand out; the wide aperture also allows a short exposure.
Only partly right. A camera focuses by moving the lens (changing $v$ ). In the eye, the lens-to-retina distance is fixed; the ciliary muscles change the shape of the lens and so its focal length (accommodation). Both change the setting until the image falls on the light-sensitive surface.

Quick check

A photographer changes the aperture from f/4 to f/8 and keeps the exposure time the same. What happens to the light reaching the sensor?

Key points

A camera lens forms a real, inverted, diminished image on a sensor; it focuses by moving the lens, and $v$ increases as the object comes closer.
f-number $N = \frac{f}{D}$ ; exposure $\propto \frac{t}{N^2}$ ; each stop (×√2 in $N$ ) halves the light.
A small aperture (large $N$ ) gives a large depth of field but needs a longer exposure.

Sources and credits

25.6: Image Formation by Lenses, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Olympus Zuiko OM 50 mm lens.jpg by Petar Milošević, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Aperture, Explained | The Mysterious f Number, Emmett Budd, embedded from YouTube.

The explanations of focusing, exposure and depth of field, the two diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1A box that remembers light

2The parts of a camera and the image it forms

3Focusing: moving the lens

4Aperture, f-number and exposure

5Experiment 1.1: a model camera

A box that remembers light

The parts of a camera and the image it forms

Focusing: moving the lens

Aperture, f-number and exposure

Experiment 1.1: a model camera