Lens/Optic

[Jon Bang]

Hello, its just me again. Today i have a question about optic/lens. I have bought a Ximea MQ013CG-E2 camera for my 8mm/S8mm filmscanner. But i have never get the idea how to get the right lens....The camera has a c-mount, and i have now a cheap chinese lens. It works, but i think with this camera i will probably need something more usefull to put the film through :-) :-)

But i see that other filmscanners with the same camera uses 50mm lens. WIll that do the job? From the filmgate to the camera is it about 8cm where its very narrow, its about 2,5cm where i can put the lens, thats why the question. I can see my english is not the best either....
Any ideas?

[Nantawat Kittiwarakul]

Assuming that the camera's sensor is more or less the same size as the super8 film, hence 1:1 magnification ratio.
If the ABSOLUTE DISTANCE from the film plane to the image sensor is 8 cm (80mm) the theorical ideal focal length should be around 20mm (the magnification at twice focal length will be 1:1)

Short answer considering 50mm lens - it won't work.

But since this is grossly/oversimplified you might try with the closest available focal length - 25 mm lens instead. Expecting some trial&error of course.

[Jon Bang]

Assuming that the camera's sensor is more or less the same size as the super8 film, hence 1:1 magnification ratio.
If the ABSOLUTE DISTANCE from the film plane to the image sensor is 8 cm (80mm) the theorical ideal focal length should be around 20mm (the magnification at twice focal length will be 1:1)

Short answer considering 50mm lens - it won't work.

But since this is grossly/oversimplified you might try with the closest available focal length - 25 mm lens instead. Expecting some trial&error of course.

Ok, so the sensor is about 10cm from the film. But just curious: Why 20mm focallength when distance is 8cm? Just trying to understand the math....
The size of a super8 is 3,79X4,01mm. The size of the sensor is: 6,9x5,5mm. Thats the active area. info from the tech-sheet. How will that math be?

[Stan Jelavic]

As Nantawat said, if the distance from the film to the sensor is 80 mm using the lens formula
1/f = (1/a + 1/b) where
f =focal length
a = distance from lens center to sensor
b = distance from lens center to film
d = a + b distance from sensor to film
Need one more thing, magnification. Using proportion triangles you get:
m = a/b

For m = 1 (sensor and film same size)

a/b = 1

Plug it into the lens formula

1/f = (1/a + 1/a) = 2/a

f = a/2

d = 2a => a = d/2

f = (d/2)/2 = d/4 = 20 mm


Now for your case:

m horizontal = 6.9/3.79 = 1.82
m vertical = 5.5/4.01 = 1.37

Use 1.37 for m so that the image fits the sensor

m = 1.37

a/b = 1.37

a + b = 100mm

1.37b + b = 100

2.37b = 100

b = (100/2.37) = 42.19

1/f = (1/a + 1/b) => f = ab/(a + b) = 1.37b**2/2.37b = (1.37/2.37)*b = 0.578*b

f = 0.578*42.19 = 14 mm

Hope this answers your question. This gives you a rough guideline.
The lens should be a good quality with macro capabilities i.e. it should be able to work at 40 mm distance to the object.
If you get the lens that does not work at this distance you will get a good focus in the center and blurry edges and potentially
other aberrations and distortions.
The F stop of the lens should be around 5. Ideally adjustable aperture is preferable.

[Jon Bang]

As Nantawat said, if the distance from the film to the sensor is 80 mm using the lens formula
1/f = (1/a + 1/b) where
f =focal length
a = distance from lens center to sensor
b = distance from lens center to film
d = a + b distance from sensor to film
Need one more thing, magnification. Using proportion triangles you get:
m = a/b

For m = 1 (sensor and film same size)

a/b = 1

Plug it into the lens formula

1/f = (1/a + 1/a) = 2/a

f = a/2

d = 2a => a = d/2

f = (d/2)/2 = d/4 = 20 mm


Now for your case:

m horizontal = 6.9/3.79 = 1.82
m vertical = 5.5/4.01 = 1.37

Use 1.37 for m so that the image fits the sensor

m = 1.37

a/b = 1.37

a + b = 100mm

1.37b + b = 100

2.37b = 100

b = (100/2.37) = 42.19

1/f = (1/a + 1/b) => f = ab/(a + b) = 1.37b**2/2.37b = (1.37/2.37)*b = 0.578*b

f = 0.578*42.19 = 14 mm

Hope this answers your question. This gives you a rough guideline.
The lens should be a good quality with macro capabilities i.e. it should be able to work at 40 mm distance to the object.
If you get the lens that does not work at this distance you will get a good focus in the center and blurry edges and potentially
other aberrations and distortions.
The F stop of the lens should be around 5. Ideally adjustable aperture is preferable.

Ok. Thanks for very good explanation...:-)

[Stan Jelavic]

Any time Joe. In practice you may have to tweak the distances somewhat.