- #1

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Any help is appreciated.

Thanks.

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- Thread starter questions_uk
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- #1

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Any help is appreciated.

Thanks.

- #2

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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

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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

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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

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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

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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

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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

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

Something that is pretty much assumed, but maybe not obvious to students, is that

See if you can use that, plus the "1/f = ..." and magnification equations you listed, to solve the problem.

- #9

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

Something that is pretty much assumed, but maybe not obvious to students, is thatMis 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

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If you knew the width/height ratio for the display, you could figure out the horizontal dimension rather than just "assume a value".

- #11

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