Is the 2X zoom on an iPhone 8 Plus digital or optical zoom

[Stanley]

Hi Bob,

OK. That was very useful. Thank you.

You have a nice website, by the way.

Stanley

[rjlittlefield]

Both of the iPhone 8+ cameras are 12MP cameras, and both share a common die size, however the wide camera has has a larger pixel pitch ( 1 vs ~1.22 um ).

I'm puzzled by the phrase "common die size".

If two sensors have the same pixel count, and one has a larger pitch than the other, then the one with larger pitch will also have a proportionally larger active area.

Does "common die size" mean that the one with smaller pitch has more wasted space around the outside? Or something else, and if so what?

--Rik

[Stanley]

Both of the iPhone 8+ cameras are 12MP cameras, and both share a common die size, however the wide camera has has a larger pixel pitch ( 1 vs ~1.22 um ) . So the ratio will not be 2:1. It should be something like 2 * ( 1 / 1.22 ) = 1.64:1 which is certainly close enough to your observed value.

-- Bob

Hi Bob or anyone else,

Besides Rik's question, I am also perplexed by something else in Bob's answer.

I went back to the two photographs that I have in Photoshop Elements -- one of them of my house with the 1X magnification and the other one of them with the 2X magnification. I looked at the horizontal and vertical rulers that I use to measure the pictures. The unit for the rulers is pixels -- not centimeters or inches, for example.

In the 1X photograph, from one point to another measures 830 pixels. In the 2X photograph, from that same point to the other same point measures 1660 pixels.

Thus, if one of the lenses has a larger pixel pitch, then the number of pixels should vary from one picture to the other. But that is not the case. The 2X picture measures exactly twice as many pixels.

Therefore, I seem to be back to square one.

Stanley

[rjlittlefield]

Stanley,

Aside from the wording about "common die size", everything makes sense to me. Let's see if I can explain...

The size of field covered by an image is proportional to sensor size divided by focal length. Double the sensor size and you double the size of the field; double the focal length and you halve the size of the field.

For the iPhone 8+, I've seen the camera described as "12 MP, f/1.8, 28mm (wide)" and "12 MP, f/2.8, 57mm (telephoto)". The 12 MP and f/1.8 are ordinary values, but the "28mm" is given in terms of what lens a 35mm fullframe camera would need to get the same size field. That type of specification is a handy way of making more clear what the lens and sensor will do instead of what they are.

But the lens focal lengths that you're seeing in the images are real focal lengths, and to make sense of those, we need to know the sensor sizes.

Working forward from Bob's ~1.22 microns per pixel number, multiplying by the number of pixels in the image (4032x3024), we conclude that the iPhone's sensor is about 4.919x3.689 mm, or about 6.149 mm diagonal.

Meanwhile, the 35mm fullframe camera has a sensor size that is 36x24 mm, or about 43.27 mm diagonal.

With those numbers in hand, we can see that the "fullframe equivalent" of a 3.99 mm lens on 6.149 mm (diagonal) sensor is 3.99 * 43.27 / 6.149 = 28.08 mm on a fullframe 43.27 mm (diagonal) sensor.

Similarly, we can compute the dimensions of the other sensor, with pitch = 1.0 microns, as being 4.032 x 3.024 mm, 5.04 mm diagonal. Then that sensor's 6.60 mm lens is equivalent to 6.60 * 43.27 / 5.04 = 56.7 mm on fullframe.

Rounding the calculated 28.08 mm and 56.7 gives us the "28 mm" and "57 mm" quoted by the source that I've linked.

In simpler math, the wide angle camera has a lens that is 1.654 times shorter, and a sensor that is 1.22 times wider, so it covers a field that is 1.654 * 1.22 ~= 2 times wider, with the same number of pixels.

--Rik

[Stanley]

Hi Rik,

Thank you so much for the great explanation. You are an amazing teacher.

Also, I am glad to learn about the website to which you linked (gsmarena).

Stanley

[Stanley]

Hi Rik (and of course anyone else),

Your explanation of pixels with my 3.99 mm iPhone 8 Plus focal-length camera really intrigued me. It got me to wondering if I could determine the actual size of an object from its image, since I do have information on the pixel values.

I was recently at a botanical garden, and I was able to get nice and close to a butterfly with my iPhone. I took the picture that I have posted, and I know that I used 1X zoom. Nothing special about the picture, but I am posting it just so that we can all see it. The length of the wings from one end to the other comes to 1460 pixels.

We do need to know the camera’s distance from the object, and that I just don’t know. But let’s say that it was, oh, about 250 mm. Once I see the formulas used to do the calculation, then I could play around with the numbers.

So here is my question, please. Given that I was using a 1.2 MP, f/1.8, 28 mm lens, with a pixel pitch of 1.22 μm, with a wing length of 1460 pixels on the image, and on the assumption that the camera was 250 mm from the object, can we determine the actual size of the butterfly?

Thank you.

Stanley

[rjlittlefield]

Given that I was using a 1.2 MP, f/1.8, 28 mm lens, with a pixel pitch of 1.22 μm, with a wing length of 1460 pixels on the image, and on the assumption that the camera was 250 mm from the object, can we determine the actual size of the butterfly?

Sure. But instead of using that "28mm" number and then having to correct for the difference between fullframe and actual sensor sizes, it's simpler to just use the real focal length of 3.99 mm.

First, let's assume that the lens focuses closer simply by moving farther away from the sensor, and see how far the lens was from sensor when the photo was taken. The relevant formula (a very good approximation) is 1/f = 1/o + 1/i, where f is the focal length, o is the distance from lens to object, and i is the distance from lens to sensor. Given f = 3.99 mm and o = 250 mm, then i ~= 4.05 mm. The lens is 4.05 mm from the sensor.

As a separate calculation, the image of the butterfly is 1460 pixels * 0.00122 mm/pixel = 1.78 mm on sensor.

After that, we just use similar triangles to argue that the width of the butterfly itself must have been 1.78 mm * (250 mm / 4.05 mm) = 109.9 mm.

Ignoring the minor change in lens-to-sensor distance based on focus, the calculated size of the butterfly will be just proportional to its distance from the lens.

Make sense?

--Rik

[Stanley]

Hi Rik,

Fantastically useful information.

Thank you so much.

Stanley

[Bob-O-Rama]

Only a portion of the die is the actual image sensor, the rest is not wasted, per se, as there are other design considerations. You still have the same needs for I/O if your pixel pitch is 6 um or 1 um. As an example, for layered designs, the size of the sensor may be largely dictated by image processor it is mounted to.

Anyway, I could not find any hard references to back up my numbers, its just something that stuck in my memory. A while back there was some site which does tear-downs and has a lot of chip porn and macro / micro work - perhaps someone here will remember what it was. I am sure that is where I read about it. I though it was "chip works" or something.

[rjlittlefield]

Yeah, "wasted" was not exactly the right term. But still, it seems unlikely that active circuitry surrounding the sensor area would naturally change to exactly compensate for the difference in pixel size. Good point about the layered designs -- that issue had not occurred to me.

I did some searching for detailed info about the sensor chips, but I couldn't find what I really wanted either. Seems like it would have to be on some tear-down site. Apple likes to talk about what their cameras do, not so much what they are.

--Rik

[Bob-O-Rama]

Found it...

https://www.techinsights.com/blog/apple ... s-teardown

They apparently absorbed the chip imaging organization I was thinking of. Sounds like a fun place to work.

[rjlittlefield]

Excellent -- thanks!

--Rik