Diffraction - Related to Effective Aperture (Pupil Size)?

[SoupOrPhoto]

When I think about diffraction limiting sharpness at higher focal ratios (f stops) in photography, I have always thought about this in terms of diffraction experiments in physics. A small slit will lead to photons magically becoming waves with resultant interference or diffraction (I know this is not how a physicist would explain it--I like substituting "magic" for anything weird relating to quantum mechanics).

It just occurred to me that I may not be thinking about this the right way. The reason I question myself is because people often speak of diffraction related to the focal ratio. I would expect that diffraction effects would actually be related to the diameter of the entrance pupil (i.e., the effective aperture).

If this were true, rather than talking about diffraction in relation to focal ratio, it should be considered in relation to the effective aperture (D).

f = (focal length, F)/(diameter of entrance pupil = effective aperture, D) = F/D

so

D (diameter of entrance pupil) = (focal length, F) / (focal ratio, f) = F/f

If diffraction is related to the physical size of the entrance pupil (D), it seems that the effects could be predicted based on D alone. So if you are using a lens with twice the focal length (100mm vs. 50mm, for example), you can double the focal ratio for equivalent diffraction effects. If diffraction effects are seen at f/11 on the 50mm lens, similar effects will be seen at f/22 on a 100mm lens.

I think I must have some fundamental misunderstanding because I have not noticed anyone speaking about diffraction this way. What am I missing?

 

[Erik Kaffehr]

When I think about diffraction limiting sharpness at higher focal ratios (f stops) in photography, I have always thought about this in terms of diffraction experiments in physics. A small slit will lead to photons magically becoming waves with resultant interference or diffraction (I know this is not how a physicist would explain it--I like substituting "magic" for anything weird relating to quantum mechanics).

It just occurred to me that I may not be thinking about this the right way. The reason I question myself is because people often speak of diffraction related to the focal ratio. I would expect that diffraction effects would actually be related to the diameter of the entrance pupil (i.e., the effective aperture).

If this were true, rather than talking about diffraction in relation to focal ratio, it should be considered in relation to the effective aperture (D).

f = (focal length, F)/(diameter of entrance pupil = effective aperture, D) = F/D

so

D (diameter of entrance pupil) = (focal length, F) / (focal ratio, f) = F/f

If diffraction is related to the physical size of the entrance pupil (D), it seems that the effects could be predicted based on D alone. So if you are using a lens with twice the focal length (100mm vs. 50mm, for example), you can double the focal ratio for equivalent diffraction effects. If diffraction effects are seen at f/11 on the 50mm lens, similar effects will be seen at f/22 on a 100mm lens.

I think I must have some fundamental misunderstanding because I have not noticed anyone speaking about diffraction this way. What am I missing?

Hi,

Think of diffraction in angular terms and you realise that it will be projected on the sensor. A longer focal length will project it over a longer distance, so it will be more enlarged.

But, you are right that the diameter of the aperture is determining angular resolution. That is the reason that astronomers that much interested in the diameter of the telescope. A telescope with twice the diameter will collect four times the light and will diffuse it over one fourth of the area. So a telescope with twice the diameter will be able to detect (or analyze light from) stars having 1 sixteenth of the brightness.

But, astronomers always discuss angular resolution and not resolution as lp/mm.

The aperture is a dimensionsless number, that happens to magically practical in many calculations.

Best regards

Erik

 

[alanr0]

When I think about diffraction limiting sharpness at higher focal ratios (f stops) in photography, I have always thought about this in terms of diffraction experiments in physics. A small slit will lead to photons magically becoming waves with resultant interference or diffraction (I know this is not how a physicist would explain it--I like substituting "magic" for anything weird relating to quantum mechanics).

It just occurred to me that I may not be thinking about this the right way. The reason I question myself is because people often speak of diffraction related to the focal ratio. I would expect that diffraction effects would actually be related to the diameter of the entrance pupil (i.e., the effective aperture).

If this were true, rather than talking about diffraction in relation to focal ratio, it should be considered in relation to the effective aperture (D).

f = (focal length, F)/(diameter of entrance pupil = effective aperture, D) = F/D

so

D (diameter of entrance pupil) = (focal length, F) / (focal ratio, f) = F/f

If diffraction is related to the physical size of the entrance pupil (D), it seems that the effects could be predicted based on D alone. So if you are using a lens with twice the focal length (100mm vs. 50mm, for example), you can double the focal ratio for equivalent diffraction effects. If diffraction effects are seen at f/11 on the 50mm lens, similar effects will be seen at f/22 on a 100mm lens.

I think I must have some fundamental misunderstanding because I have not noticed anyone speaking about diffraction this way. What am I missing?

As Erik said, diffraction by a circular aperture broadens the angular spread of light by an amount proportional to the ratio of wavelength to aperture diameter.

More specifically, the angular radius of the first dark ring of the Airy Disk is 1.22 λ / D
in radians where λ is the wavelength, and D the entrance pupil diameter.


If you are building a telescope, this will tell you the angular resolution of the instrument - assuming your optics are good enough.

For photography, we are more often interested in the amount of blur measured at the sensor. To convert angular spread to spatial broadening in the image plane, we simply multiply by the image distance, which is approximately equal to the focal length, f, of the lens.

Radius of Airy disk in image plane: r = 1.22 λ f / D = 1.22 λ F#
Which only depends on the focal ratio or f-number F#.

A 50 mm F/8 lens and a 100 mm F/16 lens will have the same 6.25 mm entrance pupil diameter and the same angular resolution (0.1 milli-radian or 22 arc seconds), but the 100 mm lens will produce a more magnified image, with twice the blur due to diffraction (10.7 μm at F/16, compared with 5.4 μm at F/8).

 

[SoupOrPhoto]

When I think about diffraction limiting sharpness at higher focal ratios (f stops) in photography, I have always thought about this in terms of diffraction experiments in physics. A small slit will lead to photons magically becoming waves with resultant interference or diffraction (I know this is not how a physicist would explain it--I like substituting "magic" for anything weird relating to quantum mechanics).

It just occurred to me that I may not be thinking about this the right way. The reason I question myself is because people often speak of diffraction related to the focal ratio. I would expect that diffraction effects would actually be related to the diameter of the entrance pupil (i.e., the effective aperture).

If this were true, rather than talking about diffraction in relation to focal ratio, it should be considered in relation to the effective aperture (D).

f = (focal length, F)/(diameter of entrance pupil = effective aperture, D) = F/D

so

D (diameter of entrance pupil) = (focal length, F) / (focal ratio, f) = F/f

If diffraction is related to the physical size of the entrance pupil (D), it seems that the effects could be predicted based on D alone. So if you are using a lens with twice the focal length (100mm vs. 50mm, for example), you can double the focal ratio for equivalent diffraction effects. If diffraction effects are seen at f/11 on the 50mm lens, similar effects will be seen at f/22 on a 100mm lens.

I think I must have some fundamental misunderstanding because I have not noticed anyone speaking about diffraction this way. What am I missing?

As Erik said, diffraction by a circular aperture broadens the angular spread of light by an amount proportional to the ratio of wavelength to aperture diameter.

More specifically, the angular radius of the first dark ring of the Airy Disk is 1.22 λ / D
in radians where λ is the wavelength, and D the entrance pupil diameter.

If you are building a telescope, this will tell you the angular resolution of the instrument - assuming your optics are good enough.

For photography, we are more often interested in the amount of blur measured at the sensor. To convert angular spread to spatial broadening in the image plane, we simply multiply by the image distance, which is approximately equal to the focal length, f, of the lens.

Radius of Airy disk in image plane: r = 1.22 λ f / D = 1.22 λ F#
Which only depends on the focal ratio or f-number F#.

A 50 mm F/8 lens and a 100 mm F/16 lens will have the same 6.25 mm entrance pupil diameter and the same angular resolution (0.1 milli-radian or 22 arc seconds), but the 100 mm lens will produce a more magnified image, with twice the blur due to diffraction (10.7 μm at F/16, compared with 5.4 μm at F/8).

Erik and Alan,

Thanks. I knew I was missing something, but was just not figuring it out. This makes sense. The focal ratio is pretty cool.

Soup

 

[Erik Kaffehr]

When I think about diffraction limiting sharpness at higher focal ratios (f stops) in photography, I have always thought about this in terms of diffraction experiments in physics. A small slit will lead to photons magically becoming waves with resultant interference or diffraction (I know this is not how a physicist would explain it--I like substituting "magic" for anything weird relating to quantum mechanics).

It just occurred to me that I may not be thinking about this the right way. The reason I question myself is because people often speak of diffraction related to the focal ratio. I would expect that diffraction effects would actually be related to the diameter of the entrance pupil (i.e., the effective aperture).

If this were true, rather than talking about diffraction in relation to focal ratio, it should be considered in relation to the effective aperture (D).

f = (focal length, F)/(diameter of entrance pupil = effective aperture, D) = F/D

so

D (diameter of entrance pupil) = (focal length, F) / (focal ratio, f) = F/f

If diffraction is related to the physical size of the entrance pupil (D), it seems that the effects could be predicted based on D alone. So if you are using a lens with twice the focal length (100mm vs. 50mm, for example), you can double the focal ratio for equivalent diffraction effects. If diffraction effects are seen at f/11 on the 50mm lens, similar effects will be seen at f/22 on a 100mm lens.

I think I must have some fundamental misunderstanding because I have not noticed anyone speaking about diffraction this way. What am I missing?

As Erik said, diffraction by a circular aperture broadens the angular spread of light by an amount proportional to the ratio of wavelength to aperture diameter.

More specifically, the angular radius of the first dark ring of the Airy Disk is 1.22 λ / D
in radians where λ is the wavelength, and D the entrance pupil diameter.

If you are building a telescope, this will tell you the angular resolution of the instrument - assuming your optics are good enough.

For photography, we are more often interested in the amount of blur measured at the sensor. To convert angular spread to spatial broadening in the image plane, we simply multiply by the image distance, which is approximately equal to the focal length, f, of the lens.

Radius of Airy disk in image plane: r = 1.22 λ f / D = 1.22 λ F#
Which only depends on the focal ratio or f-number F#.

A 50 mm F/8 lens and a 100 mm F/16 lens will have the same 6.25 mm entrance pupil diameter and the same angular resolution (0.1 milli-radian or 22 arc seconds), but the 100 mm lens will produce a more magnified image, with twice the blur due to diffraction (10.7 μm at F/16, compared with 5.4 μm at F/8).

Erik and Alan,

Thanks. I knew I was missing something, but was just not figuring it out. This makes sense. The focal ratio is pretty cool.

Soup

On the other side,

Assume that the 50 mm lens will cover a 43 mm image circle with good sharpness. Converting the same design to 100 mm would cover an image circle 86 mm. So that lens would yield four times the information, if image circle would be fully utilized.

But, that is not normally the case. Lenses are mostly optimized to cover the intended image circle.

Best regards

Erik

 

[Great Bustard]

When I think about diffraction limiting sharpness at higher focal ratios (f stops) in photography, I have always thought about this in terms of diffraction experiments in physics. A small slit will lead to photons magically becoming waves with resultant interference or diffraction (I know this is not how a physicist would explain it--I like substituting "magic" for anything weird relating to quantum mechanics).

It just occurred to me that I may not be thinking about this the right way. The reason I question myself is because people often speak of diffraction related to the focal ratio. I would expect that diffraction effects would actually be related to the diameter of the entrance pupil (i.e., the effective aperture).

If this were true, rather than talking about diffraction in relation to focal ratio, it should be considered in relation to the effective aperture (D).

f = (focal length, F)/(diameter of entrance pupil = effective aperture, D) = F/D

so

D (diameter of entrance pupil) = (focal length, F) / (focal ratio, f) = F/f

If diffraction is related to the physical size of the entrance pupil (D), it seems that the effects could be predicted based on D alone. So if you are using a lens with twice the focal length (100mm vs. 50mm, for example), you can double the focal ratio for equivalent diffraction effects. If diffraction effects are seen at f/11 on the 50mm lens, similar effects will be seen at f/22 on a 100mm lens.

I think I must have some fundamental misunderstanding because I have not noticed anyone speaking about diffraction this way. What am I missing?

The effect of diffraction softening is a result of the spread of light due to diffraction. The amount of spread is proportional to the [reciprocal of] the relative aperture (quotient of focal length and the diameter of the effective aperture -- entrance pupil).

However, it is not the absolute spread of the light that matters, but rather the relative spread of that light as a proportion of the linear dimension of the sensor, which is crucial in cross-format comparisons.

For example, while f/16 on FF will spread the light twice as much as f/16 on mFT, the spread will subtend the have the proportion on a FF sensor as an mFT sensor, because the FF sensor has twice the linear dimension. Thus the *effect* of the diffraction softening will be twice as much for f/16 on mFT as f/16 on FF.

Let's work an example with numbers:

• f/16 on FF: for green light (550 nm), the diameter of the Airy Disk will be 2.44 x 550 nm x 16 = 21.5 μm. 21.5 μm / 43.3mm = 0.05% the diagonal of the sensor.
• f/16 on mFT: for green light (550 nm), the diameter of the Airy Disk will still be 2.44 x 550 nm x 16 = 21.5 μm, but 21.5 μm / 21.6mm = 0.1% the diagonal of the sensor.

Thus, we see that while the absolute amount of blur is the same, the relative blur (which is what matters with regards to the blurring of the photo) is twice as much on mFT as on FF. Hence, for cross-format comparisons, we would use the diameter of the effective aperture (entrance pupil) rather than the relative aperture to compare the effects diffraction.

 

[57even]

The amount of blur you see at the sensor is a product of the aperture that created the blur and the magnification of that blur at the sensor.

In other words, the image of a dot is a blurred disk. For a given aperture, the size of the blurred disk at the sensor will be larger the more the image is magnified, and magnification depends on focal length.

So, you have to multiply the blur, which is inversely proportional to aperture size D, by the magnification, which is proportional to focal length f.

Blur is proportional to f/D, which is f/number.