Light Transmission Through Lenses

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This article collates information previously published together with new information to provide a complete guide to factors affecting how light from a scene finally arrives at the camera sensor. In addition, it points out the pitfalls in using too simplistic an approach if light levels are critical.


The lens aperture, introduction

The f-number of a lens is the ratio of the focal length to the effective object lens diameter. It is a mechanical ratio and does not infer the efficiency of a lens. It does affect the amount of light energy passed to the sensor and will play a significant part in the resulting picture. Traditionally camera manufacturers have specified sensitivity with a lens having an aperture of f 1.4. This would be fine if they all did it the same but they don’t. Some say with 75% reflectance some say 89% and so on. Then again, some will state the sensitivity with AGC on but not what the AGC gain is. Camera specmanship is too vast a subject to expand on in this article but suffice to say the f-number of the lens is a most important consideration. In simple terms the smaller the f-number the more light is passed to the sensor, therefore f1.2 is better than f1.8. The percentage of light passed by different apertures is listed in table 1. This shows the percentage of light falling on the lens that is passed to the sensor.

F number f1.0 f1.2 f1.4 f1.7 f2.0 f2.8 f4.0 f5.6
% passed 20% 14.14% 10% 7.07% 5% 2.5% 1.25% 0.625%

Table 1 light passed by f-stops

Yes, it is true that with an aperture of f1.4 only 10% of the light on the lens is passed to the sensor. Some manufacturers specify camera sensitivity as that on the faceplate or sensor. In these cases use these ratios to convert to the light required on the lens. I.e. 1-lux faceplate sensitivity requires 10 lux at the scene with an f-1.4 lens, or 20 lux with an f-2.0 lens.

It may seem relatively unimportant to quibble about the difference between an f-1.2 lens and an f-1.4 lens, especially when the latter is much cheaper than the first. It is significant though because the f1.4 lens needs 50% more light for the same energy on the sensor. An example taken from distributors’ catalogue shows two 12mm auto iris lenses, one f1.4, and one f1.2. The f1.4 is £75.00 the f1.2 is £152.00! Which lens would a comparatively inexperienced estimator use in a competitive tender? In a ten-camera system, it could mean a saving of £750.00 or about £1,000.00 on the tender price- a considerable temptation.

How aperture numbers are calculated

The scale of ‘f’ numbers, 1.4, 2.0, 2.8, 4, 5.6, etc. is such that successive numbers halve the amount of light passed to the sensor. These particular numbers are known as "full stops" and are in the ratio of Cons3.gif. This only applies to "full stops"; there are also half stops, which are numbers half way between full stops, and one-third stops In other words, the amount of light is proportional to the cross sectional area of the light rays entering the lens..

It can be shown that the f number,

Cons4.gif

From this equation the following can be derived;

Cons5.gif

If the illuminance is defined as ‘E’ lux, the equation can be shortened to;

Cons6.gif

If reflectance (R) is taken into account, this now becomes,

Cons7.gif

Another consideration is, how efficient is the lens at passing the maximum amount of light. There are several factors that determine the efficiency of a lens and of course they all cost money. When light passes through a glass/air boundary some is lost through reflection and refraction; this is reduced in the more expensive lenses. In addition, different light frequencies are refracted at different angles; special coatings are used to ensure that all frequencies are transmitted in parallel rays. The factor that measures the efficiency of a lens is known as the transparency ratio, (t) from which is calculated the transmission ratio (T).

The transparency ratio (t) is a function of the lens design and the number of glass elements. This is generally only available from the manufacturer. The transmission ratio (T) is the effective lens stop after adjusting for the transparency ratio (t) and is defined by; Cons8.gif. The resulting number will always be larger than the specified f-number.

For example, an f-1.4 lens having a transparency ratio (t) of 0.785 would have an effective aperture of f 1.58. This would be the value to use when calculating the light transmitted through a lens rather than the published f-number. All reputable manufacturers should be able to provide information on the transmission ratios for their lenses.

If this is now taken into account, the relationship becomes.

Cons9.gif

If an example is now worked using;

Transparancy ratio, t=0.785,

Reflectance, R=0.89,

Scene illumination, Escene=15 lux

Lens aperture, f=1.4,

Then, the light required on the sensor, Cons10.gif

For those who wish to go into a little more depth, the transparency ratio (t) is derived from the formula;

Cons11.gif

Where; t = transparency ratio

g = Reflectance ratio, decided by the maker, typically 0.015

n = the number of lens surfaces (one lens element has two surfaces).

Therefore for an f1.4 lens with 16 lens surfaces and g =0.015, the value of ‘t’ will be, 0.785


Summary

The critical factor that determines camera performance is the amount of light reaching the sensor and it can be seen that there are many potential losses to be allowed for. The only accurate way to calculate the light level required or estimate camera performance in given conditions is to use the light required by the sensor and work back through the formulae. Formula (5) can be transformed as;

Cons12.gif


The effect of sensor size

There is yet another factor that affects the efficiency of a camera/lens combination.

As stated in a previous article, light is energy measured in Watts per square Metre. Therefore, if the area of a sensor is known then the resultant power in watts can be calculated. The nominal areas of the sensors in common use are listed in table 2.

Table5.gif

Table 2, areas of sensors

The power produced by each individual pixel in the sensor is directly proportional to its area. If three cameras are considered each with the same resolution of say 500 lines then the number of pixels on each sensor must be the same. The result of this is that the pixels on each smaller size of sensor must also be smaller. Therefore, the power produced will be less for the same aperture setting, i.e. the same amount of light energy. It is assumed that the light energy to produce a full 1-volt pp video signal is 5.0 mW/M2

If for the sake of an example, a light source of 1,000 milliwatts per square Metre is passed to the sensor via an f-1.8 aperture lens. The amount of light passed by the lens will be 7.5% = 75 mW/M2. From this, the power output can be calculated for each sensor. This will be the power multiplied by the area of the sensor. The result is shown in table 3.

Table6.gif

Table 3-power output of sensors

Therefore the 1/2" and 2/3" sensors will be producing insufficient power for a full video signal. The answer is to use a lens with a larger aperture for these sensors so that more energy is passed to maintain the output power. This is summarised in table 4.

Table7.gif

Table 4-power output corrected by lens f-stop

This is the reason that many 1/3" cameras have the sensitivity specified with an f-1.0 or sometimes an f 0.9 aperture. Beware though, there are only a limited number of lenses made to the 1/3" format. If the longer focal length lenses must be used they usually have smaller apertures (higher f-numbers) and pass less light energy.

The contra to this argument is that if a sensor of one size has the same size pixels as a larger one, then the light required will be the same. However the total number of pixels will be fewer and the resulting resolution will be proportionally less. There is no such thing as a free lunch!


Sensor and lens format

The range of lenses available for the 1/2" and particularly the 1/3" cameras is limited. The largest range of lenses is still 2/3" and 1" format. It is all right to use a larger format lens on a smaller format camera but there is another penalty to pay. A 2/3" lens will focus the scene on the equivalent area around a 1/2" sensor but only the light energy relative to the area of the sensor will be converted to power. See diagram 1.

2/3" lens with a 1/2" sensor

The area of a 1/2" sensor is 53% that of a 2/3" sensor, therefore only this proportion of the light energy will be converted to output power and thus a video signal. This would make an f-1.4 lens equivalent to about f-2.0. The area of a 1/3" sensor is only 25% that of a 2/3" which is equivalent to an f-2.8 lens! Therefore, this is yet another correction factor that needs to be applied.

Conclusion

The f-number of a lens and as listed in a camera specification is an important guide to the eventual usefulness of a system. However, if a system is being designed where lighting is critical and the camera is being expected to perform at the printed specification you could finish up with disappointing results. A cheap lens with a poor transmission ratio could need 50% more light than a more expensive lens with an apparently similar specification. A 2/3", f-1.4 lens fitted to a ½" camera will pass the equivalent amount of light to an f-2.0 lens. Again, take care with some inexpensive zoom lenses, the f-number may vary with the focal length.

A not untypical scenario could be as follows:

A lens is listed in distributors' brochure as being 12:1 zoom, 2/3", F-1.8. An enquiry to the manufacturer elicits the information that the transmission ratio is 0.65 and the aperture at maximum zoom is f-2.4. You plan to use the lens on a ½" camera to its specification of 1.0 lux at f-1.4.

Using the data in this article, it can be shown that the effective aperture when zoomed, will actually be equivalent to f-5.0! The minimum scene illumination required would be 12.5 lux, twelve times the camera specification! No doubt, everyone would put the blame on the camera and although I strongly attack camera manufacturers for the manner in which their specifications are presented, in this case it would be unjust.

The adjustments for colour cameras can be very critical to performance. For instance if the results for a monochrome camera with a sensitivity of 1.0 lux were out by a factor of three, it would mean a difference of three lux. In the case of a colour camera with a sensitivity of 7.0 lux, the difference would be 21.0 lux, a significant variation.


Summary

The factors that will affect the ultimate amount of light energy reaching a camera sensor are summarised below.

  1. The lens f-number.
  2. The f-number changing with zoom setting.
  3. The transmission efficiency of the lens.
  4. The lens size compared to the sensor size.
  5. The scene reflectance.
  6. Spectral response.


Using the table

A table of f-stops and the associated attenuation to light is attached. The columns are:

Short list of full and half stops.

A) The attenuation to light passed calculated from the f-number

B) The reciprocal of ‘A’, The factor by which the light entering the lens must be multiplied by, to calculate the light at the sensor.

C) ‘B’ expressed as a percentage of light passed to the sensor.

Note that these values are applied to the f number as a mechanical ratio. They need to be adjusted by the factors previously discussed.

The values in column ‘A’ may be used to compare camera specifications on a like-for-like basis.

E.g., Camera. ‘a’, 1 lux at f 1.4,

Camera. ‘b’, 0.7 lux at f 0.7

The performance for Camera. ‘b’ at f1.4 would be; Cons14.gif lux, or......

The performance for Camera. ‘a’ at f 0.7 would be; Cons15.gif lux

Another use for the table is if a lens is necessary, that has smaller minimum aperture than the camera specification, i.e. a long focal length zoom lens.

E.g., camera specified as 3 lux at f 1.2 is using a lens with an aperture of f 2.8

The light required would be; Cons16.gif

In the absence of a camera manufacturer providing the light required on the sensor, it can be estimated by multiplying the light level given at a particular aperture by the associated factor in column ‘B’,

I.e. 2 lux at f 1.2 would be;

2 x 0.1389 = 0.28 lux at the sensor.

Formula (6) may then be used to calculate the minimum scene illuminance required. Using faceplate illumination and lens data will always produce a more reliable answer for scene illumination.

If you are not sure which numbers to use to multiply or divide by, use simple logic. For instance, if you are calculating the scene illumination for a larger f-number you will expect a higher value.

One final point is that, just like cameras, lenses have a spectral response diagram. This needs to be checked especially if infrared illumination is being used.

Table8.gif

(*) f-1.8 is actually a 1/3 stop but is included because many lenses have this specification.

(**) These numbers are rounded off, the true value is shown in brackets.


You can make up a simple spreadsheet to calculate formula (6) as follows.

Put the following headings in the columns;

Table9.gif

Into cell E2, type the following;

=(A2*B2^2)/(C2*D2*0.2)

Enter the values into cells A2, B2, C2 and D2, the result will show under E Scene in cell E2.

Table10.gif

If you wish, you can calculate the sensor illuminance by entering the following formula into cell A3;

=(C2*D2*E2*0.2)/(B2^2)

Then enter the values in cells B2, C2, D2 & E2, the result will be displayed incell A3. Note, before any values have been entered, cells E2 and A3 may show "divide by zero" error.

This chapter is supplied by Mike Constant and was originally published in CCTV Today. Mike is the author of 'The Principles & Practice of CCTV' which is generally accepted as the benchmark for CCTV installation in the UK.

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