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I'm gonna introduce my concrete example, but I'd like to know how do do this calculation in general. Say I have a middle-format camera with film height 58mm and a 90mm/3.5 lens. What is the equivalent aperture on a 24mm film with 50mm lens that has the same circle of confusion/DoF?

Edit for clarification: Suppose I have two cameras with lenses. One has frame height of 24mm and lens with focal length of 50mm. The second camera has frame height 56mm and a lens focal length of 50mm full frame equivalent (which should be around 117mm physical, I think). Now suppose I take a photo with the full frame camera with its aperture wide open at let's say f/1.8. A subject at focal plane will be sharp and a background at some distnce will be blurred to a disc of size which can be represented as a fraction of a frame. Now suppose I take the exact scene with the medium format camera. But if I open the aperture to f/1.8 on that one, the blur disc of the background will be relatively larger. So what should I step my aperture down to to have the same relative blur size.

In even simpler terms, how do I calculate bokeh equivalency? If I have some bokeh on a medium camera at some particular fstop, what would be the fstop to have an equivalent bokeh on a full frame, ceteris paribus?

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  • \$\begingroup\$ What is your planned display size? Viewed from what distance? Ultimately, all of these variables play a part in what CoC you need to use to get the DoF you desire. \$\endgroup\$
    – Michael C
    Jul 26, 2020 at 8:13

3 Answers 3

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Calculation of the equivalent lens range

Classically, to calculate the equivalent lens focal range \$f\$, you need to use the diagonals (\$D_1\$, \$D_2\$) of the two films/sensors to calculate the 'crop factor' \$k = D_1/D_2\$.

The equivalent lens focal range is then given by the formula: \$f_2 = f_1/k\$

In your case (from the clarification part of your question):

For 24x36mm film/sensor, \$D_1 = \sqrt{24^2+36^2} = 43.27\text{ mm}\$

For 6x9 films, \$D_2 = \sqrt{56^2+84^2} = 100.96\text{ mm}\$

Hence \$k = 0.429\$

So for a 6x9, the equivalent focal range to a the 50mm lens on the 24x36, is \$50/0.429 = 116.67\text{ mm}\$. This confirm your own calculation.

Calculation of the equivalent aperture

With the risk of simplification(*) : source Wikipedia:

$$ \mathrm{DoF}(N,D) = \frac{2NcD^2}{f^2} $$ with:

  • \$\mathrm{DoF}\$: depth of field
  • \$N\$: aperture
  • \$c\$: circle of confusion in mm.
  • \$D\$: distance to the subject (focus distance)
  • \$f\$: focal length in mm

At the equivalent aperture, if it exists, the spots of the dots on both films must have diameters (\$c_1\$, \$c_2\$) such as \$c_2 = c_1 / k\$.

So, when \$N_1 = k N_2\$,

$$\begin{align} \mathrm{DoF_2}(N,D) &= \frac{2 N_2 c_2 D^2}{f_2^2} = \frac{2 N_2 c_1/k D^2}{(f_1/k)^2} \\ &= \frac{2kN_2 c_1 D^2}{f_1^2} \\ &= \mathrm{DoF_1}(k N_2,D) \\ &= \mathrm{DoF_1}(N_1,D) \end{align}$$

This confirms, on the one hand, that there is an equivalent aperture, and gives us the formula to calculate it.

Formula giving the equivalent aperture \$N_2 = N_1/k\$

In your case

  • \$N_1 = 1.8\$ and \$k = 0.429\$ so the equivalent aperture is \$1.8 /0.429 = \mathbf{4.2}\$

Note: (*) \$D\$ needs to be much greater than \$f\$.

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    \$\begingroup\$ Yes. This is the calculation I had in my mind. I didn't know if it was really a function of "only" the crop factor or if there were some other variables which would make the final calculation/formula much more difficult. \$\endgroup\$ Aug 4, 2020 at 13:23
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As to exposure, the lens aperture is unchanged. Your lens is always f/3.5, it is never affected by film size. It is affected by very close macro focus distances (fstop Number = focal length / aperture diameter).

As to Field of View, any Equivalent focal length is computed from the diagonal, so it matters if your film is 6x6 cm, 6x7 cm, or 6x9 cm. Assuming 6x6 cm, the diagonal is 79.2 mm (which is crop factor 0.55x).

The 35 mm diagonal is 43.3 mm, so the field of view on 35 mm film would be the same if using a (90 mm x 43.3/79.2) = 49.2 mm focal length on 35 mm film. My site has that calculator at https://www.scantips.com/lights/cropfactor3.html (third calculator).

My site also has a DOF calculator at https://www.scantips.com/lights/dof.html which will allow insisting on any CoC you care to enter. That does not change the DOF that the lens and film does, but it does change the resulting calculated DOF distance numbers (where that CoC would apply).

Circle of Confusion is affected by print enlargement, which is affected by film size (if printing the same standard 8x10 print). On 35 mm film, it would change the standard 6x6 CoC of 0.055 mm to become 0.03 mm for 35 mm film, which is the same 43.3/79.2 factor.

But standard CoC (as viewed enlarged to the standard 8x10 inch print size) is computed from (diagonal mm / a divisor), which divisor is normally either 1500 or 1442. It is often said to be 1500, but the standard CoC numbers we see are usually computed with 1442. CoC is not really optional, it relates CoC to the smallest point the human eye might see on an 8x10 inch standard enlargement (but human eyesight does vary).

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The job of the camera lens is to handle each point on the subject (or vista) and project that point on film or digital sensor as a point of light. This projected point of light is the smallest fraction of an image that carries intelligence. The image thus consists of countless such points of light. These show up as tiny circles, juxtaposed with neighboring circles and each has diffused boundaries. We call them “circles of confusion”.

When we view a displayed pictorial image, should it consist of circles of confusion that are too tiny to be resolved, we declare that image (or portion of an image) to be in-focus. This needed circle size is not engraved in stone. It is a variable based on viewing distance, image brightness, image contrast, image content, and the visual acuity of the observer.

What is the diameter of the largest circle of confusion that appears as an unresolved point to an observer? An object like a coin is viewed as a disk at close range. As the observer recedes from the coin at a distance of about 3000 diameters, the coin is seen as a point and not a disk. This is a quite a small ratio. This would be like a 3 foot diameter wagon wheel seen from 1.7 miles. However, this criterion is far too stringent for pictorial photography, because of the contrast limitations of our media. A reduced specification of 3.4 minutes of arc is the accepted value. This works out to a circle viewed from 1000 times its diameter. This is 1/100 of an inch circle viewed from 10 inches or 2/100 of an inch viewed from standard reading distant of 20 inches. In the metric system this works out to 0.5mm viewed from standard reading distance.

Now the format size we call 35mm full frame = 24mm height by 36mm length. This is a miniature image worthless unless somehow magnified for viewing. Suppose you make a display image 8 by 12 inches. To accomplish with exact cropping, you apply 8.5 X (magnification). Say the image appears tack sharp. To accomplish, the circles of confusion must be 0.5mm in diameter of smaller. This dimension is assigned to the final displayed image. At the image plane of the camera, the permissible circle size is 0.5 ÷ 8.5 = 0.06mm.

A larger format camera, say format size 60mm by 90mm, will require less magnification to make a 8 x 12 inch display image. For this larger film size we need only apply 3.4 X (magnification). The circle size to do this deed is 0.5 ÷ 3.4 = 0.15mm at the film plane.

Bottom line – Depth-of-field tables are only a best guess: it is ridiculous to believe a table that says 4 feet 9 inches is spot on.

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