Am I right on this?

[Phazan]

Ok, so I think I just figured out something...

If I have a 50mm lens, and the apeture is say, at 2.8. I do 50 divided by 2.8 and it gives me 17.85. So does that mean that the opening is 17.85mm from top to bottom?
That would explain why 300mm 2.8 lenses are so expensive, because when you do 300/2.8 you get 107, which is a pretty big hole, and you have to have that much more glass to cover it...And the reason why when you have a more shallow DOF with a telephoto lens, is because the hole on a 2.8 tele lens is HUGE, compared to a 17mm F2.8 lens with a 6mm hole.
And then the reason why I can't get very shallow DOF on my P&S camera is because it's a 6-72mm lens..so the hole turns out to be about 2mm in diameter!

Is this true?? If it is, I'm so proud of myself for figuring that out...Hope it wasn't too confusing

 

[json2001]

Yeah pretty sure thats accurate math.

f#=(Focal Length)/Diameter

so...

Diameter=F#/(Focal Length)

As to whether or not that is the reason telephotos are more expensive...I don't really know, but your hypothesis makes a whole lot of sense to me!

 

[prodigy2k7]

bump...we need more replies!

 

[KhronoS]

Yeah pretty sure thats accurate math.

f#=(Focal Length)/Diameter

so...

Diameter=F#/(Focal Length)

As to whether or not that is the reason telephotos are more expensive...I don't really know, but your hypothesis makes a whole lot of sense to me!

Diameter=(Focal Length)/F# this is correct

@ Phazan, yeah you are pretty correct with your assumptions

 

[Phazan]

Ok, awesome! Thanks for answering!

 

[Steph]

Previous answers are not entirely accurate. The true definition for the F number is: F=focal length/diameter of entrance pupil.The entrance pupil is the image of the physical aperture as seen through the front element of the lens.

 

[prodigy2k7]

something about this doesnt sound right...

I meant on the math part...

300mm/F2 = 150mm

so

150mm * F2 = 300mm

 

[Big Mike]

Remember that the Aperture value is actually a fraction or ratio...so when we say F2.8...it's actually 1:2.8 or 1/2.8...so you would multiply the aperture value by the focal length to get the diameter of the entrance pupil (or divide by the denominator)...so you have the right idea.

And Yes, that is a big reason why F2.8 zoom lenses are bigger and more expensive than F4 zoom lenses and a lot more expensive than F4-5.6 lenses.

 

[prodigy2k7]

Remember that the Aperture value is actually a fraction or ratio...so when we say F2.8...it's actually 1:2.8 or 1/2.8...so you would multiply the aperture value by the focal length to get the diameter of the entrance pupil (or divide by the denominator)...so you have the right idea.

And Yes, that is a big reason why F2.8 zoom lenses are bigger and more expensive than F4 zoom lenses and a lot more expensive than F4-5.6 lenses.

Well I think he is just comparing the same aperture lenses with different focal lengths...

But ya, I said F to designate that its the aperture

 

[Alex_B]

Well I think he is just comparing the same aperture lenses with different focal lengths...

But ya, I said F to designate that its the aperture

f is the focal length.

therefore an aperture of f/N means an aperture which is the focal f length divided by N. N is sometimes also referred to as f-number or the effective aperture.

So this should explain what apertures f/2.8 and f/8 and so on mean.

 

[prodigy2k7]

f is the focal length.

therefore an aperture of f/2.8 means an aperture which is the focal f length divided by 2.8. 2.8 is sometimes also referred to as the effective aperture.

oh ok thanks im dumb ha..

So...back to what the OP said, hes right, right?

 

[Alex_B]

oh ok thanks im dumb ha..

So...back to what the OP said, hes right, right?

i think he is.

however, when it comes to DoF on P&S cameras, the situation is not as simple, since also the sensor is much smaller. but in a first approximation he is right.