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Focal length

  • #1
Hi. It asks to calculate the focal length range of a projector of square pixel SXGA is from 4 to 12 m from the screen. It gives the diagonal size of the display (in mm) as well as the horizontal size (in m). The thing that am not sure about is how the focal length would be calculated with the screen dimensions and whether info about square pixels and SXGA would be required. I assume the range means to calculate the focal length at 12 then at 4 and take these values away from one another?

Any help is appreciated.

Thanks.
 

Answers and Replies

  • #2
Redbelly98
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I am guessing you'll need to take the magnification into account, which means you would need to know the size of the screen as well.

The equation for magnification, and other stuff about lenses, is here:
https://www.physicsforums.com/library.php?do=view_item&itemid=148

The question sounds a little weird, since the focal length of a lens is not adjustable.
 
  • #3
I am guessing you'll need to take the magnification into account, which means you would need to know the size of the screen as well.

The equation for magnification, and other stuff about lenses, is here:
https://www.physicsforums.com/library.php?do=view_item&itemid=148

The question sounds a little weird, since the focal length of a lens is not adjustable.

Thanks for the reply. Interestingly enough I also wondered whether the magnification would be taken into the account, as magnification is included in the notes. What was a bit weird was how it gave the diagonal dimension of the display and the horizontal dimension of the screen?
 
  • #4
Redbelly98
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What was a bit weird was how it gave the diagonal dimension of the display and the horizontal dimension of the screen?
Okay. Looks like you'll have to figure out the horizontal dimension of the display, in order to get the magnification.
 
  • #5
Okay. Looks like you'll have to figure out the horizontal dimension of the display, in order to get the magnification.
That's what I thought, and then compare the display with the screen. However, the magnification depends on U and V which I think are the lengths from the display to the lens and from the lens to the screen.
 
  • #6
Redbelly98
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That's what I thought, and then compare the display with the screen. However, the magnification depends on U and V which I think are the lengths from the display to the lens and from the lens to the screen.
Yes, that's right.

It would be useful, if you haven't done so already, to list the equations that are relevant to solving this problem.
 
  • #7
Yes, that's right.

It would be useful, if you haven't done so already, to list the equations that are relevant to solving this problem.
Hi. Some of the equations I've got include:

1 / f = 1 / v + 1 / u

And M = v / u

Also am not sure if this is related to it but

Numerical Aperture / F number

N = F / D

Thanks!
 
  • #8
Redbelly98
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Okay, good. I'm thinking the numerical aperture and lens diameter won't come into play here, at least I haven't seen anything yet that would relate to them. (Of course, I also haven't seen the actual problem statement either.)

The magnification equation and "1/f = ..." are important here.

Something that is pretty much assumed, but maybe not obvious to students, is that M is also, by definition,

M = (width of screen) / (width of display)​

See if you can use that, plus the "1/f = ..." and magnification equations you listed, to solve the problem.
 
  • #9
Okay, good. I'm thinking the numerical aperture and lens diameter won't come into play here, at least I haven't seen anything yet that would relate to them. (Of course, I also haven't seen the actual problem statement either.)

The magnification equation and "1/f = ..." are important here.

Something that is pretty much assumed, but maybe not obvious to students, is that M is also, by definition,

M = (width of screen) / (width of display)​

See if you can use that, plus the "1/f = ..." and magnification equations you listed, to solve the problem.
Thanks for your help! I basically used the other magnification rule you gave and assumed a value for the horizontal dimension.

This could then be be used with M = v / u thus calculating u as v and M are already known and obtaining a v value. With the u and f values I used 1 / f = 1 / u + 1 / v to obtain the focal length at 4 and 12 m. Then obtained the difference between these two values to obtain the range.

But yeah I did assume a value for the horizontal component. It gives the diagonal dimensions.

Thanks!
 
  • #10
Redbelly98
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Sounds like you pretty much have it then.

If you knew the width/height ratio for the display, you could figure out the horizontal dimension rather than just "assume a value".
 
  • #11
Thanks for your help. I used the resolution of the dosplay to work out the horizontal dimensions of it.
 

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