This is the size at which the subject is reproduced on the sensor i.e. if the subject is 36mm long and it's image on the sensor chip is 36mm long as shown in the image of an engineers steel rule, taken on a 5D with a 36mm sensor chip, the magnification is 1x or 1:1


Actual size 1:1
on sensor/film chip
5x Enlargement of image on this screen of chip.
36mm © Phil Gee
  5x © Phil Gee

Mag =image size /subject size

Magnification is independent of format if you use a full frame 36 x 24mm camera (Canon 5D) or a 22 x 14mm (Canon 40D)camera or even a 5" x 4" sheet film camera.
A 22mm subject at 1:1 will be recorded at 22mm on each film/sensor chip, there may be problems getting it all on the smaller sensor but it's still 1:1.



Image on Chip
36:7 © Phil Gee

The rule was again photographed so that 7mm (subject) fills the frame of the 36mm wide chip giving an image 36mm wide then the magnification is:

M = 36/7 = 5.14x 

you can also determine the magnification using 

focal length (nominal value marked on lens) of the lens used 

Mag = extension /focal length 
with a 50mm lens and 50mm of extension 

M = 50/50 = 1 or 1x or 1:1 

If the lens is moved further away from the sensor by adding another 50mm extension tube you now have 100mm of extension and the magnification will be:

M = 100/50 = 2 or 2x i.e. the subject will appear twice as big on the sensor

Moving the lens further away from the sensor means that the light is spread over a much larger area so the amount of light falling on the sensor is reduced by a proportionate amount (inverse square law) which effects exposure,


Exposure and Magnification 

By how much do you need to increase the standard exposure value to allow for this reduction in light due to magnification.

Photograph of rule at lenses closest focus point so within normal exposure limits no correction required
220 mm recorded on 36mm chip = 36/220 = 0.164x
220mm on 36 © Phil Gee


Portion of Magnified image recorded on 36mm sensor chip

7mm on 36 © Phil Gee

As can be seen above for normal exposure values to be used all 220mm of the image would need to fall on the sensor as only a fraction does in the magnified portion (7mm) more exposure needs to be given to compensate this can be calculated as follows:

Exposure factor (Exf)= (1+Mag)^2

for a 2x magnification the exposure needs increasing by

Exf = (1+2)^2 = 3^2 = 3x3 = 9

Nine times the normal exposure is required to compensate

which is equivalent to 3 and a bit stops some cameras have halve or third stop intervals to accommodate intermediate values.

1 stop = ± 2x as much light 
2 stop = ± 4x
3 stop = ± 8x
4 stop = ± 16x  
5 stop = ± 32x
6 stop = ± 64x

First you need to calculate the Magnification

In the case of the steel rule example above 7mm of the rule is covering the full with of the sensor which is 36mm wide.

Magnification = 36/7 = 5.14x.

Exf = (1+ 5.14)^2 = 6.14^2 = 37.6996 or 38x which is just over 5stops

Using a SLR you can easily determine magnification you need to know the size of the sensor image you see in the viewfinder if it is full frame approx 36mm
so if you place a rule at the point of focus and read off the width shown; if you are using a large format you can use a rule on the ground glass screen as a comparison.

Lets say you are using a 35mm SLR and can see 72mm of the rule at the subject position the magnification is:

M= 36/72 = 0.5x 
i.e. the image of the subject is half real size on the sensor

Exf = (1+0.5)^2 = 1.5^2 = 1.5x1.5 = 2.25 times more light

if you can only see 16mm then

M = 36/16 = 2.25 x
(exposure compensation Exf = (1+2.25)^2 = 3.25|2 =10.56 =3.5 stops)

However if you wanted it to be exactly 2x magnification you would have to adjust the set up until you could see how many mm in sharp focus     _ _ mm ?


Mag   Exf   Stops
0.1x 1.21 1/3
0.2x  1.44 1/2
0.3x 1.69 2/3
0.4x 1.96 1
0.5x 2.25 1.25
1x 4 2
2x  9 3.33
3x 16 4
4x  25 4.66
5x  36 5.33

Now you should have a rough idea of how much extra exposure each magnification value requires (in stops).
Using this information how does it help to find the correct exposure? 

First you need to determine the standard exposure required at the subject by either rule of thumb = BSD rule or using an exposure meter,
or if you have a camera with a through the lens meter (TTLM) you could check it using a standard lens.
You may wonder why bother if you have TTLM then that will give you a corrected reading taking into account the extension - correct
But say you are using a Medium or large format camera which has no TTLM. 
A 35mm SLR in addition to giving you a reference shot can act as a exposure spot-meter! in-fact many large format landscape photographers use a Digital SLR to give a reference image and light meter readings.
Once you have the standard exposure reading for the subject apply the correction by manipulating either the Aperture or Shutter speed or both depending on the result you are trying to achieve.

If the standard exposure for subject is ƒ16 at 1/100 sec

You are reproducing a 6mm subject to 18 mm on the sensor,  apply correction to the standard exposure and show the results if you applied it only to: 

1. Aperture setting 
2. Shutter speed 
3. A combination of both


Depth of Field and Magnification

I prefer to use the term Zone of Critical Focus rather than DoF as at high degrees of magnification you are working at mm and sub mm levels so precise focusing and knowledge of the limits of the zone that will reproduce as sharp is critical.

The smaller (higher the magnification) you go the narrower becomes the Depth of Field (DoF) or (ZoCF) when you are working at these high degrees of magnification the limits of the ZoCF are equally disposed about the PPF.

You need to remember that the degree of enlargement to the final viewing size is an important factor as is the final viewing distance if the image is to appear sharp when viewed. ref Circle of Confusion (CoC) notes


As a rough guide using a aperture of ƒ16  

Mag ZoCF @ƒ16
1:1 (1x) 2.24mm
2:1 (2x) 0.84 mm
3:1 (3x) 0.498 mm 
4:1 (4x) 0.35 mm 
5:1 (5x) 0.269 mm