Photo PhysicsVersion 060519
INTRODUCTION
This resource is under development.

CONTENTS









Focal plane
This is the plane where the image is formed by the lens of a camera.
It is where the image sensor (film or digital) is located.
It is usually indicated by the symbol on the top of the camera.
If the focal plane is not marked, then it can be assumed that it is within 10mm of the rear of the camera.

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Aperture
The aperture controls the amount of light passing through a lens.
The aperture, f number or stop number is equal to the ratio of the focal length of the lens to the aperture diameter.
f number = focal length / aperture diameter,
where the focal length and aperture diameter are measured in the same units, e.g. mm
The aperture setting also determines the Depth of Field obtained in an image as well as the effects of Spherical Aberation from the lens.

Aperture calculator
To find f number
Enter:-
the focal length mm,
the aperture diameter mm,
and then press
f number =
To find focal length
Enter:-
the aperture f number
the aperture diameter mm,
and then press
focal length = mm.
To find aperture diameter
Enter:-
the focal length mm,
the aperture f number
and then press
aperture diameter = mm.

E.g.1   The eye has a focal length of approximately 22mm and a lens aperture in the dark of approximately 4mm.
Therefore, the eye has an equivalent f number of f5.5.
E.g.2   A reflecting telescope with a 150mm mirror and a focal length of 500mm has an aperture f number of f3.3

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Light
Light is a very small part of the electromagnetic spectrum with wavelengths between approximately 700nm (red) and 450nm (violet).
All wavelengths of light (and of the electromagnetic spectrum) travels in straight lines at a speed of approximately 300,000,000 metres per second in a vacuum.
The speed of light in a vacuum is now a universal constant and is defined as 299,792,854m/s.
It is usually represented by the symbol c.

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Thin Lens Formula
The lenses on all modern cameras are made up of several thinner lenses to provide focussing and zoom facilities as well as to correct for spherical and chromatic aberation.
Although they are not 'thin' lenses, the following formula is useful for understanding the relationship between the distance of the object from the lens (u), the distance of the image formed from the lens (v) and the focal length of the lens (f).
All of these distances must be in the same units e.g. mm.
The thin lens formula is 1/u + 1/v = 1/f

Real is positive sign convention
In order to be able to use the thin lens formula, a 'real is positive sign convention' is used.
In this convention, 'real' is when rays of light actually pass through the object, image or focus. These values will be positive.
When rays of light only appear to pass through the object, image of focus, then these are said to be 'virtual'. The values for these will be negative.
Three calculators are given below for when the object distance, image distance or focal length are unknown.

Thin lens calculator
Enter:-
To find object distance
Enter:-
focal length (f) mm,
image distance (v) mm,
press
object distance (u) = mm
To find image distance
Enter:-
focal length (f) mm,
object distance (u) mm,
press
image distance (v) = mm
To find focal length
Enter:-
object distance (u) mm,
image distance (v) mm,
press
focal length (f) = mm

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Extension tubes
Extension tubes are hollow tubes fitted between the lens and camera body to enable the camera to focus closer than normal.
In doing so, they enable the lens to produce a larger image on the camera sensor, so enabling close up and macro photography.
Extension tubes can be used with prime and zooms, wide angle and telephoto lenses, but it is worth considering how the extension tube will alter the focussing range before starting.

Extension tube calculator
The term 'closest distance from camera' in the calculator below refers to the nearest distance that can be focused with the lens before the addition of extension tubes.
This should be measured from the focal plane of the camera.
Enter:-
focal length of lens mm,
closest distance from camera mm,
length of the extension tube mm,
and then press
New focus range from focal plane = mm to mm
At the closest focussing position, the magnification increases from to


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Linear Magnification
The diagram below shows an image of height i formed by a lens of an object of height o.

Linear magnification is defined as the height of the image divided by the height of the object.
Linear magnification = i/o
The image height and the object height are difficult to measure, but the object height is proportional to the object distance and the image height is proportional to the image distance.
{tan α = o/u = i/v}
Therefore Linear magnification = v/u

Linear Magnification calculator.
Enter:-
object distance (u) mm,
image distance (v) mm,
and then press
Linear Magnification =

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Digital Sensor Sizes
The list shows the approximate physical size of the sensors used on digital cameras.
(Full frame is the 35mm film equivalent.   APS-C is the Advanced Photo System Classic film equivalent.)
The actual size of the sensor depends on the manufacturer.
The number of pixels on a sensor is not determined by its physical size.
E.g. for a full frame sensor, the Nikon D4S has 16.2 Mpixels while the Nikon D850 has 47.5Mpixels.
Sensor name
1/2.3 inch
1/1.7 inch
1 inch
Four Thirds
APS-C
APS-H
Full frame
Medium format
 Approximate size (mm)
6.3 x 4.7
7.6 x 5.7
13.2 x 8.8
17.3 x 13
23.5 x 15.6
26.6 x 17.9
36 x 24
44 x 33
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Field of View
The field of view is the angular range that can be seen.
For humans, the field of view for binocular vision is approximately 120°.
However, our visual perception is mainly based on the central 50° or so of our vision.

When applied to cameras, the field of view will depend on the distance of the lens from the sensor and the size of the sensor.

{Looking at the diagram above, the field of view = 2 × tan-1(s/2v)}

Field of View Calculator.
Enter:-
size of sensor (s) mm,
image distance (v) mm,
and then press
Field of View =

Example 1.
Consider a camera with a full frame sensor (36mm) fitted with a 35mm focal length lens.
When focussed on a distant object the image distance will be 35mm and so the landscape field of view is 54.4°.

Example 2
Consider a camera with an APS-C sensor (23.5mm) fitted with a 35mm focal length lens.
When focussed on a distant object the image distance will be 35mm but the landscape field of view is only 37.1°.

To have this field of view with a full frame sensor would require a lens with a focal length of 53.5mm, (1.53 times larger).
Therefore the smaller the sensor, the smaller the focal length of the lens to achieve an equivalent field of view to a full frame camera.

Some lenses are manufactured specifically for smaller sensors and so will have restricted fields of view.
E.g. the Nikon AF-S DX Nikkor 35mm f1.8G lens, which only has a maximum field of view of 44°.
This will work fine for an APS-C sensor camera (37.1°), but will produce a restricted image on a full frame sensor (54.4°)

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Ray Tracing
A ray of light is a very narrow beam of light.
It will travel in a straight line unless it passes from one material to another.
Light from an object passing through a lens will form an image.
Three rays of light are needed to find where the image is formed.
Consider the diagram below.

Any rays of light passing through the optical centre of a thin lens will continue in a straight line.
A ray of light travelling parallel to the principal axis will be refracted by the lens to pass through the focus of the lens (marked F).

The principal axis represents a ray of light travelling from the bottom of the object, through the centre of the lens. It will continue in a straight line onto where the image is formed.
The ray of light shown red, travels from the top of the object, to the centre of the lens. It will continue in a straight line onto where the image is formed.
The ray of light shown in green leaves the object and travels parallel to the principal axis. It is refracted by the lens to pass through the focus of the lens and then onto where the image is formed.
The image is then formed where the rays, shown green and red, cross.
All rays of light are reversible, and the ray of light shown blue represents a ray of light leaving the object and passing through the focus of the lens. It continues onto the lens where it is refracted to travel parallel to the principal axis and then onto the image.

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Converging lens - image position
The following ray diagrams show the relationship between the position of the object and the image for a thin converging lens.
Object at infinity.

The image is formed at the focus (F) of the lens. It is inverted, small and real.
This situation is used in refracting telescopes and photography.

Object between infinity and twice the focal length of the lens (2F).

The image is formed between F and 2F. It is inverted, small and real.
This situation is used in most photography.

Object is at twice the focal length of the lens (2F).

The image is formed at 2F. It is inverted, the same size and real.
This situation is used in most macro photography and photocopiers.

Object is between 2F the focus (F) of the lens.

The image is formed beyond 2F. It is inverted, bigger and real.
This situation is used in most macro photography and projectors.

Object is at the focal length of the lens.

The image is formed at infinity. It is inverted, bigger and real.
This situation is used in search lights, torches etc.

Object is closer than the focal length of the lens.

No real image is formed. Where the rays appear to cross an upright, bigger and virtual image is seen.
This situation is used in magnifying glasses.


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Power of a lens
The power of a lens is defined as the reciprocal of the focal length of the lens when measured in metres.
The power of a lens is measured in Dioptres (D) and is positive for a converging lens and negative for a diverging lens.
A converging lens with a focal length of 100mm (0.1m) will therefore have a power of 1/0.1 = 10D.

Lens Power Calculator.
Remember - for a converging lens the focal length is positive - for a diverging lens the focal length if negative. Enter:-
the focal length of the lens mm,
and then press
Power of Lens = D


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Refraction
When light travels from one substance to a different substance, there is usually a change in its speed.
E.g. The speed of light in air is approximately 300,000,000m/s (3 × 108ms-1)
In glass, the speed of light is approximately 200,000,000m/s and in water approximately 225,000,000m/s.
The ratio of the speed of light in a vacuum to the speed of light in a substance is called the 'Refractive Index' of the substance and is represented by the letter n.

Therefore refractive index of a substance = velocity of light in a vacuum / velocity of light in a substance (v).
nsubstance = c/v

Refractive index is a ratio of velocities and so has no units.
E.g.   nair ≈ 1.0003,   nglass ≈ 1.5,   nwater ≈ 1.33

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When light changes speed at the boundary between two substances, some of the light is reflected and some is transmitted into the new substance. The amount of reflected and refracted light depends on the two substances.
When the light is incident at right angles to the boundary, the transmitted light continues in the same direction, and the reflected light travels back along its original path.

When the light is incident at an angles to the boundary, the reflected light is reflected at an equal angle to the incident light, and the transmitted light changes direction.
If the refractive index of substance 1 is less than that of substance 2, then the refracted light is bent towards the normal.

If the refractive index of substance 1 is greater than that of substance 2, then the refracted light is bent away from the normal.


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Ray incident from the left.

Ray incident from the left.


Snell's Law
What is now known as 'Snell's Law' was first accurately described by a Persian scientist, Ibn Sahl, in 984. He is reported to have used his law to derive the shape of lenses that focus light without geometrical aberration.
Many other scientists over the years produced alternative derivations including Williebrord Snellius, a Dutch astronomer, in the 17th century.
The mathematical version of Snell's Law is:-
  velocity of light in substance 1 = sin (angle of incidence)
  velocity of light in substance 2 = sin (angle of refraction)

This becomes   n1 × sin (i) = n2 × sin (r)
where:-
n1 is the refractive index of substance 1 and
n2 is the refractive index of substance 2.


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Thin Lens - types
A 'Thin lens' is a single piece of refracting material.
Most photographic lenses consist of multiple thin lenses.
The shape of a lens determines how the light passing through will be refracted.
There are two main types of thin lenses - CONVERGING and DIVERGING.
A Converging lens causes parallel rays of light to a bend together, while a Diverging lens causes parallel rays of light to bend apart.

Converging lenses are always thicker at their centre. This can be achieved in three main ways.

Diverging lenses are always thinner at their centre. This can be achieved in three main ways.


Passage of parallel rays of light through thin lenses. Rays enter the lenses from the left.
   

 

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Diverging Lens - image position
The following ray diagrams show the relationship between the position of the object and the image for a thin diverging lens.
Object beyond F.

The image formed for any object position beyond F is upright, smaller and virtual.
This situation is mostly used to correct for short sight in spectacles and contact lenses.

Object at F.

The image formed when the object is at F is upright, half of the size and virtual.

Object between F and the lens.

The image formed is upright, larger than half of the size and virtual.


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Visual acuity
Visual acuity is a measure of the spatial resolution of the visual system.
Spatial resolution is measured in minutes of arc.   (1° containing 60 minutes)
Visual acuity = 1 / gap size (arc min.)
For 'Normal vision', visual acuity =1, so for normal vision, the gap size is 1 arc min.
This corresponds to 20/20 = 1 (imperial), or 6/6 = 1 (metric) vision as indicated by eye tests.
(Many people can resolve to around 0.4 arc min.)
This is also affected by the Diffraction of light through the iris of the eye.

Greatest visual definition is normally when the object is held 250 - 300mm from the eye.
(1 arc min. = 0.01667°   = 2.91 × 10-4 radians)
When an object is viewed at 250mm, the smallest distance that can be resolved is the arc length of 1 arc min.
Therefore smallest distance that can be resolved = 250 × 2.91 × 10-4 = 7.27 × 10-2mm
Any details smaller than this will not be resolved by 'normal' eyes.

This has several important consequences for:-
Printer resolution,
Monitor resolution,
Depth of field.


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Printer resolution
(See Visual acuity.) Printers often provide the option to select the print resolution.
Common resolutions include 150, 200, 300, 600 and 1200 dots per inch (dpi).
A 'normal' eye, looking at a well lit object 250mm away, will just be able to resolve details which are approximately 0.073mm apart.
Most modern printers work by putting tiny dots of ink or toner onto the paper.
So if the dots are closer together than 0.073mm, then the eyes will not be able to resolove these individual dots and they will appear to merge together.
0.073mm corresponds to 1/0.073 = 13.7 dots per mm.
With 25.4mm per inch, this corresponds to 348 dots per inch (dpi).
The dots produced by inkjet type printers are often less well defined that those of laser printers, as the ink will 'bleed' into the paper.
Therefore, selecting 300 dpi on an inkjet printer will often just produce a clear image, while 600 dpi will probably be necessary with a laser printer.
In practice, it is worth using 600 dpi for both laser and inkjet printers - 1200dpi being unnecessary.

Although most modern picture editing programs will do their best to produce the best possible print from the image file available, it is worth considering how large an image file needs to be for the size of print required.
This is done by multiplying the printer resolution by the size of print required.
E.g. To print a picture at 12 × 8 inches, with a printer resolution of 600 dpi, will require an image file
that is 12 × 600 by 8 × 600 pixels = 7200 by 4800 pixels, i.e a 34.6Mpixel image!
Fortunately, most picture editing programs will interpolate between pixels for smaller image files and so a successful print can be obtained with half (or even a quarter) of this number of pixels.
However, using a file smaller than 1800 by 1200 pixels is likely to produce a picture which consists of small blocks rather than dots and will appear 'pixellated'.


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Monitor resolution
(See Visual acuity.) Most monitors and televisions (projectors) use LCD or LED technology and are to the High Definition (HD) standard which includes an aspect ratio (horizontal to vertical ratio) of 16:9.

The first HD standard was 720p, which has 720 pixels vertically.
(The p refers to a progressive scan rather than an i which was an interlaced scan used on some old CRT monitors.)
With a 16:9 aspect ratio, there are 16 × 720 / 9 horizontal pixels = 1280 horizontal pixels.
So 720p is a screen 1280 × 720 pixels.

The next HD standard was 1080p, with a screen resolution of 1920 × 1080 pixels.
(Some manufacturers introduced variants of this standard, 1920 × 1200 pixels and 2048 × variable number pixels.)

The current standard for larger monitors is the 4K standard, where the 4K refers to the horizontal instead of the vertical resolution.
4K screens either have resolutions of 4096 × 2304 (or a variable number of pixels) and are known as 'Cinema 4K' or a resolution of 3840 × 2160 pixels and are known as UHD (Ultra High Definition).

Most monitors used for editing photographs are viewed at a distance of 400 - 600mm.
A 'normal' eye has a visual acuity of 1 arc minute. (2.91 × 10-4 radians)
Taking the viewing distance of the monitor as 500mm, then the eye will just detect a distance of
500 × 2.91 × 10-4 = 0.15mm.
Taking this distance as the horizontal pixel spacing for a 1080p HD monitor,
the maximum horizontal size for the screen is 0.15 × 1980 = 297mm. (Approx 12 inches.)
By a similar calculation, the maximum vertical size is 0.15 × 1080 = 162mm.
Since screen sizes are usually given as the diagonal distance across the screen, this corresponds to a diagonal distance of √(2972 + 1622) = 338mm (13.3 inches).

A 4K monitor viewed at 500mm, the maximum horizontal screen size would be 0.15 × 4096 = 614mm and the vertical size = 0.15 × 2304 = 346mm.
This gives a diagonal distance of 705mm (28 inches).

The monitor being used for picture editing should therefore not be too large in order to prevent the image appearing as individual pixels.

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Exposure
The image sensor (film or semiconductor) of a camera needs to receive a certain amount of energy from the incident light to be activated.
The amount of energy needed depends on the film speed or sensitivity setting and is usually expressed as an ISO number, e.g. 200.
The amount of incident light reaching the sensor depends on the Intensity of the light (how bright it is), the area of the lens aperture and the time that the shutter is open for (shutter speed).

Light intensity is measured in Watts per square metre (Wm-2).
Since Energy = Power × time,
the light energy incident on the sensor = Intensity (I) × Area of lens (A) × time the shutter is open (t).
For a circular lens, area = π × radius2 = π × (diameter/2)2.

The Aperture of a lens is measured as an f number and the f number is proportional to 1/diameter of the lens.
Therefore the area of the lens is proportional to 1/(f number)2.
If the sensor sensitivity and the light intensity are constant, then A × t must also be a constant.
Therefore     the time the shutter is open     must also be constant,
                             (f number)2 ,
i.e. the time the shutter is open is proportional to (f number)2.
E.g. If an image exposes correctly with a shutter time of 0.002s (1/500s) and an aperture of f8, then a shutter time of 0.001s (1/1000s) needs an aperture of √(82/2) = f5.66.
Similarly if an image exposes correctly with a shutter time of 0.002s (1/500s) and an aperture of f8, then a shutter time of 0.004s (1/250s) needs an aperture of √(82×2) = f11.3.

The table below shows the standard f numbers and their squared values and can be used to determine what f number should be used if the time the shutter is open is halved or doubled.
f number1.422.845.7811.31622.632
(f number)22481632641282565121024

Most modern lenses also allow intermediate f numbers, e.g. the squared value half way between f2.8 (8) and f4 (16) is 12.   The square root of 12 is f3.5 - which is often the largest aperture on 'kit zoom lenses'.

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Astro exposures
Taking photographs of the night sky has three difficulties associated with ensuring sufficient light energy reaches the camera sensor.
1).   Use a large aperture lens.
A wide aperture ensures more light can reach the sensor and reduces the effects of diffraction.
Wide aperture, wide angle lenses (e.g. Nikon 35mm f1.8) are relatively cheap (£150) but become exponentially more expensive as focal length increases (e.g. Nikon 300mm f2.8, £5800).
By comparison, a 500mm focal length reflecting telescope with a diameter of 150mm only has an aperture of f3.3.

2).   Increase the sensitivity of the camera sensor.
When the eye has become accustomed to the dark, it can be considered to have an equivalent sensitivity to approximately ISO 1000.
The fastest commonly available film is Ilford Delta 3200 black and white (ISO 3200) at around £9 for 36 exposures.
However, the faster the film speed the more 'noise' is present in the image.
The same is also true for digital sensors.   While many digital cameras boast sensor speeds of ISO 100000 and higher, the amount of noise present in the image is usually considerable.

3).   Increase the exposure time.
If the shutter is open for too long, images of stars become lines rather than points of light (star trails).
The earth rotates through 360° in 24 hours.
This is equivalent to an angular speed of 360/(24×60×60) = 0.00417° per second = 0.25 arc minutes per second.
The angular speed in radians per second = 2×π/(24×60×60) = 7.27×10-5.
If f is the focal length of the lens (in mm) and t is the time the shutter is open (in s), the length of the star trail on the sensor will be f×t×7.27×10-5mm.
The final image will be viewed many times larger than the actual sensor size, by the eye at a distance of 250mm.
If the final image is M times larger than the image sensor, then the length of the star trail on the final image will be M×f×t×7.27×10-5mm.
(See Visual Acuity.)
A normal eye, viewing an image at 250mm, can just resolve a distance of 7.27×10-2mm.   So the star trail on the final image must be less than 7.27×10-2mm.
Therefore, 7.27×10-2 < M×f×t×7.27×10-5
Therefore, simplifying and rearranging, t = 1000/M×f seconds.

E.g. If the focal length of the lens is 100mm with a full frame camera (sensor 36mm) and the final image is 360mm (Magnification M=10), then t = 1000/10×100 = 1s.
So the maximum exposure time would be 1s.


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Depth of Field
There is only one position where a lens will form a sharp image of an object.
However, since a normal eye can only resolve to 1 arc minute, there will be a small range of object distances where the image on a sensor/screen will appear to be sharp.
See section on Visual Acuity.

In the diagram above, the lens produces a sharp image of the black point object onto the sensor.
Since the red point object is further away, the sharp image will be produced infront of the sensor and a circle will be produced on the sensor.
Similarly, the sharp image of the blue point object would be produced behind the sensor, and so a circle will be produced on the sensor.

The smallest detail the eye can resolve at 250mm is 7.27×10-2mm.
So if the circles produced on the sensor are smaller than this, the eye will consider them to be sharp.
Images are not usually viewed at sensor size, but many times larger.
For the eye to still consider them to be sharp, the size of the circles on the sensor must be correspondingly smaller.
If the final viewed image is ten times larger than the image sensor, then the circles on the image sensor will need to be ten times smaller, i.e. 7.27×10-3mm.
The magnification factor is the number of times larger the final viewed image is than the image sensor and has been initially set to 10 in the calculator below. This can be changed as necessary.
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Depth of Field Calculator.
The theory for these calculations
All distances are measured and given to the focal plane of the camera.
If this is not known, then use the distance from the rear of the camera.
Enter:-
the focal length of the lens mm,
the object distance from focal plane mm,
the aperture f number ,
the magnification factor ,
and then press
Depth of Field from mm to mm,



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Diffraction
(This section is under development.)
Whenever waves encounter a barrier, they will spread into the 'shadow' region behind the barrier.
This is known as Diffraction.
Diffraction effects can easily be demonstrated with water waves.
The diagram below shows plane (straight) water waves hitting a barrier.


Note how the waves spread into the shadow of the barrier.

When plane waves encounter a gap in a barrier, the waves not only diffract into the shadow of the barrier but also interfer with each other as they pass through the gap.

Directly in front of the gap, the waves are of large amplitude.
On either side there is then a region where the waves cancel, and then reappear weaker, and then cancel etc.
The mathematics to represent diffraction of plane waves at a gap are 'interesting'.
It can be shown that the intensity of the waves is proportional to (sin2ψ) / ψ2,
where ψ = π×d×θ/λ
(θ is the angle from the centre of the gap, d is the width of the gap and λ is the wavelength of the light.)

Light from a red laser passing through a narrow slit.
   
(Need a light picture here)
When the gap is extended to three dimensions and becomes a hole, then the effect on light becomes




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