# Does anyone know the formula for using 2/3 lens on 1/2 cdd?

### #1

Posted 11 July 2008 - 01:17 PM

### #2

Posted 11 July 2008 - 01:25 PM

-- J.S.

### #3

Posted 11 July 2008 - 02:44 PM

2/3 is 11mm diagonal and the 1/2 is 8mm, if this is any help.

Use your high school algebra to derive the formula.

### #4

Posted 11 July 2008 - 03:40 PM

Use your high school algebra to derive the formula.

The Sony EX3 is not out and still in beta tesing before release. If using a 2/3-inch type lenses with the EX3 camcorder, the Fujinon ACm-21 lens mount adapter must be used. Then your 2/3 inch lens will mount to the camera. Ref. www.sony.com/xdcamex

### #5

Posted 11 July 2008 - 03:58 PM

The focal length of a lens

*does not change*when you put it on a different size sensor or piece of film.

### #6

Posted 12 July 2008 - 01:07 AM

Recite this sentence:

The focal length of a lensdoes not changewhen you put it on a different size sensor or piece of film.

I beg to differ unless the adapter in question repositions the the focal plane to the correct size. Correct?

Recite this sentence..

"The zoom ratio is stated as being for instance 6:1 this means that the longest focal length is six times that of the shortest. The usual way of describing a zoom lens is by the format size, zoom ratio and the shortest and longest focal lengths, i.e. 2/3," 6:1, 12.5mm to 75mm. Again, great care must be taken in establishing both the camera and the lens format. The lens just described would have those focal lengths on a 2/3" camera but a range of 8mm to 48mm on a 1/2" camera. Similarly a lens giving the same performance on a 1/2" camera would be a 1/2," 6:1, 8mm to 48mm."

### #7

Posted 12 July 2008 - 03:15 AM

I beg to differ unless the adapter in question repositions the the focal plane to the correct size. Correct?

Recite this sentence..

"The zoom ratio is stated as being for instance 6:1 this means that the longest focal length is six times that of the shortest. The usual way of describing a zoom lens is by the format size, zoom ratio and the shortest and longest focal lengths, i.e. 2/3," 6:1, 12.5mm to 75mm. Again, great care must be taken in establishing both the camera and the lens format. The lens just described would have those focal lengths on a 2/3" camera but a range of 8mm to 48mm on a 1/2" camera. Similarly a lens giving the same performance on a 1/2" camera would be a 1/2," 6:1, 8mm to 48mm."

Hi,

Too much marketing hype is confusing you here. The angle of view of the lens will change but the focal length is always the same.

Stephen

### #8

Posted 12 July 2008 - 04:10 AM

I beg to differ unless the adapter in question repositions the the focal plane to the correct size. Correct?

Recite this sentence..

"The zoom ratio is stated as being for instance 6:1 this means that the longest focal length is six times that of the shortest. The usual way of describing a zoom lens is by the format size, zoom ratio and the shortest and longest focal lengths, i.e. 2/3," 6:1, 12.5mm to 75mm. Again, great care must be taken in establishing both the camera and the lens format. The lens just described would have those focal lengths on a 2/3" camera but a range of 8mm to 48mm on a 1/2" camera. Similarly a lens giving the same performance on a 1/2" camera would be a 1/2," 6:1, 8mm to 48mm."

You're buying into a lot of "effective focal length" bullshit. The focal length of a lens is a physical attribute of the lens. It does not know what it is projecting an image on so how could be possibly change focal length when you put it in front of any given format of sensor or film.

As Stephen says, you are considering the angle of view of the lens. That does change because you are using more or less of a lens' coverage by using the lens on various formats.

Take this example:

A 120mm lens on an 8x10 camera is a very wide lens. That camera/lens pair would have an horizontal angular field of view of 93 degrees.

Put that same 120mm lens on a 35mm camera (or, alternately, think of it as masking everything except a 24mmx36mm rectangle of that 8x10 groundglass) and that camera/lens pair would have an horizontal angular field of view of 17 degrees.

Notice that the focal length remained constant even though the same lens

*functioned*as a wide lens than as a long lens.

### #9

Posted 12 July 2008 - 09:54 AM

**Edited by Andrew McCarrick, 12 July 2008 - 09:57 AM.**

### #10

Posted 12 July 2008 - 10:02 AM

**Edited by Andrew McCarrick, 12 July 2008 - 10:04 AM.**

### #11

Posted 12 July 2008 - 02:43 PM

If I figured this out right you should multiple the Focal Length of the 1/2" lens by 1.375 to get the 2/3" focal length to achieve the same field of view. Or is it the inverse of that (.72)?

If you put the 2'3" lens on the 1/2" camera, the focal length will remain the same.

A 6:1, 12.5mm to 75mm lens will still be a 6:1, 12.5mm to 75mm lens.

### #12

Posted 12 July 2008 - 05:29 PM

If I figured this out right you should multiple the Focal Length of the 1/2" lens by 1.375 to get the 2/3" focal length to achieve the same field of view. Or is it the inverse of that (.72)?

Andrew, you've stated it correctly. If you have a focal length on a 1/2 inch camera and want to know what focal length will look the same on a 2/3 inch camera, you would multiply the known focal length by 1.375.

You would multiply by .72 to go the other way (know a focal length on 2/3" and want to know a focal length that will look the same on 1/2")

I will say it once again for posterity. This math above is to find equal fields of view. The focal length of each lens never changes simply from using it on a different format.

### #13

Posted 12 July 2008 - 10:09 PM

Andrew, you've stated it correctly. If you have a focal length on a 1/2 inch camera and want to know what focal length will look the same on a 2/3 inch camera, you would multiply the known focal length by 1.375.

You would multiply by .72 to go the other way (know a focal length on 2/3" and want to know a focal length that will look the same on 1/2")

I will say it once again for posterity. This math above is to find equal fields of view. The focal length of each lens never changes simply from using it on a different format.

Thanks, thats exactly what I was looking to find. "multiply the known focal length by 1.375."

### #14

Posted 12 July 2008 - 11:47 PM

Andrew, you've stated it correctly. If you have a focal length on a 1/2 inch camera and want to know what focal length will look the same on a 2/3 inch camera, you would multiply the known focal length by 1.375.

You would multiply by .72 to go the other way (know a focal length on 2/3" and want to know a focal length that will look the same on 1/2")

I will say it once again for posterity. This math above is to find equal fields of view. The focal length of each lens never changes simply from using it on a different format.

I got slightly different results, 1.32 and .75 depending on direction. I double checked but maybe I made the same mistake twice.

How did you do the math?

### #15

Posted 13 July 2008 - 01:39 AM

I got slightly different results, 1.32 and .75 depending on direction. I double checked but maybe I made the same mistake twice.

How did you do the math?

From what I found it said 2/3" is 11mm diagonal and 1/2" is 8mm diagonal, so I divided 11 by 8 and got 1.375

### #16

Posted 13 July 2008 - 01:47 AM

From what I found it said 2/3" is 11mm diagonal and 1/2" is 8mm diagonal, so I divided 11 by 8 and got 1.375

Cool, thank you.

### #17

Posted 13 July 2008 - 03:05 AM

Cool, thank you.

Hi,

I dont see how this helps in any way, if you use a lens designed for a 1/2 inch sensor it's milimeter markings will exactly mark the milimeter markings of the lens designed for the 2/3 inch sensor.

Stephen

### #18

Posted 13 July 2008 - 09:47 AM

How did you do the math?

I didn't. I took the figures stated as correct. It was the concept I had issue with.

All you do to get the two factors is divide 8 by 11 and then use 11 by 8 to go the other way.

### #19

Posted 13 July 2008 - 12:43 PM

I didn't. I took the figures stated as correct. It was the concept I had issue with.

All you do to get the two factors is divide 8 by 11 and then use 11 by 8 to go the other way.

I think that this clarifies an issue that confuses a lot of people and drives a lot of other people crazy because

they are right when they say that 'focal length is focal length' no matter what. The people who are confused probably

haven't had as much experience in different formats and so they find out that a 50mm lens seems to give a different

sized image in two different formats and focus on the focal length.

This is a useful formula for people who want to know "what lens do you use in X format to get

the closeup that a 100mm would give you in Y format?"

I should have quoted Andrew, Chris, as I was asking about his original math. I see what I did now though, thanks everybody.

### #20 Glen Alexander

Posted 13 July 2008 - 01:44 PM

1/f = 1/d1 +1/d2

M=-d2/d1