Let's talk the Physics behind the lens, shall we ?
Ok, I have an APS-C sensor camera, and I have a 50mm lens. My 50mm lens is an old full frame (FF) lens. Meaning, it's made and used before the digital sensor camera was introduced into the market.
I know that APS-C sensors are smaller than the FF sensor. How small? I'm an engineer, hence I read numbers to comprehend facts. The size of a FF sensor is 35.9mm by 24.0mm so to give an area of 861.6 sq. mm. And the size of an APS-C sensor is 23.5mm by 15.6mm giving an area of 366.6 sq.mm. Thus, APS-C sensor is in fact smaller than the FF sensor.
Hmm... sound stupid I am, right ? LOL. Well, I have time to kill, hence, just talk some crap ! LOL
Anyway, when we talked about mounting an FF 50mm lens onto the APS-C DSRL camera, the correct focal length is no more 50mm, but to multiply a crop factor of the APS-C sensor. Hence, in my case, my 50mm lens is actually a 75mm lens on my trusty A700 camera.
Sound complicated, eh ?
Well, again, I'm engineer and I speak in numbers. To understand more about this focal length change from 50mm to 75mm, I turn to my STPM Physics!
Oh crap, I had returned all my physics back to my teacher after I left school about 10 years ago. Refer back to the text book is a wise move to refresh my deep buried knowledge of optical physics.
This Advance Level Physics textbook written by Nelkon&Parker is damn boring if compared to the University Physics textbook. Well, what could I expect from a textbook that is written and published by a Brit ? LOL
Ok, turn to the page where the textbook talk about the physics of optic. There is a thin lens topic with the the Lens Equation is being derived. Not to bored myself, I skip the derivation part, and straight search for the thin lens equation, that is
The Lens Equation !
1/v + 1/u = 1/f
where
v = distance of object from lens axis,
u = distance of image from lens axis and
f = focal length of the lens.
So, I have my equation. Then how to show that a 75mm FF lens is actually 50mm APS-C lens?
Not to complicate the whole thing, I assume my lens is a single lens only. Hence, my focal length,
f is 50mm.
Say I have an object of 600mm in height. Say to cast the image onto a FF sensor, the size of the image will be 24mm. Hence, the magnification factor,
m = 24/600 = 0.04.
By definition, the Magnification Factor,
m is
v/u or size of image/size of object.
Therefore,
m = 0.04 = v/u
v = 0.04u
So, in the lens equation, rearrange, I got this
1/0.04u + 1/u = 1/50
Solve it, I got the distance of object is
u = 1,300 mm.
These numbers say : when an object of 600mm in height is placed at 1,300mm away from a 50mm lens, the image size is 24mm on a FF sensor.
So, what happen if the same object is cast onto an APS-C size sensor?
The size of the APS-C size sensor is 15.6mm. Hence, the magnification factor is 15.6/600 = 0.026.
Since the focal length of the lens doesn't change, hence the
f = 50mm. Solving the equations similar to the above give this
1/0.026u + 1/u = 1/50
u = 1,973.07mm
So, this number tells me, if I put my object at 1,973.07 mm away from the lens, the image size will be at 15.6mm on a FF sensor. But, the actual object is actually placed at 1,300mm away from the lens. Hence, the
actual lens focal length that give the image size of 15.6mm is
v = 0.04 x 1,973.07 = 78.92 mm
1/1,973.07 + 1/78.92 = 1/f
f = 75.89 mm
Hence, as conclude, when a 50mm FF lens mounted on an APS-C camera, I actually get a 75mm zoom lens instead.
So, do you understand what I wrote so far ?
I admit it sounds confusing. To sum up everything, when we mount a FF lens onto a APS-C camera, we get
extra zoom by 1.5x factor.
Got it ?
No? Never mind, because I started to confuse also ! LOL