Science and Technology Quiz 3
- Thread starter Aramazd
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I'll give it to RiffRaff, since he was the first to note Ben Franklin. But ParadigmShifter did nail the curiousity.
RiffRaff's up!
RiffRaff's up!
ParadigmShifter
Random Nonsense Generator
I'll ask another maths question then.
If you have a chessboard and cut out 2 opposite corner squares, can you cover the dissected board with 2x1 dominoes (overlapping not allowed). If you can, draw a diagram. If not, give a proof.
If you have a chessboard and cut out 2 opposite corner squares, can you cover the dissected board with 2x1 dominoes (overlapping not allowed). If you can, draw a diagram. If not, give a proof.
GoodGame
Red, White, & Blue, baby!
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Clarification, please: Are you saying that each domino is equal in dimensions to 2 squares on the chessboard by 1 square?
Also, by no overlapping, do you mean the dominos can't overlap each other, or do you mean that the dominos have to fit neatly over the disected board without overhanging? In the second case, it'd be impossible, but it'd be a good illustration of a Riemann sum.
Also, by no overlapping, do you mean the dominos can't overlap each other, or do you mean that the dominos have to fit neatly over the disected board without overhanging? In the second case, it'd be impossible, but it'd be a good illustration of a Riemann sum.
It is not possible.
Proof: The opposing corner squares are always of the same color. So that means, there will always be 30 squares of one color and 32 squares of the other color.
The two fields covered by 2x1 dominoes will always be of different color, because no fields of the same color are next to each other.
That means, no matter how you place 30 dominoes on the board, the left over squares are always of the same color, thus they cannot be next to each other and can not be covered by the 31st dominoe.
ParadigmShifter
Random Nonsense Generator
Correct, you are up next.
Suppose you make an image of an object by collimating (i.e. making a beam where the rays are parallel) the light coming from the object and then focusing the beam again on the image plane.
What happens with the image (beside obviously getting darker) , when you put a disk in the center of the collimated beam that:
a) has a smaller diameter than the beam?
b) has a bigger diameter, but a hole in the middle?
What happens with the image (beside obviously getting darker) , when you put a disk in the center of the collimated beam that:
a) has a smaller diameter than the beam?
b) has a bigger diameter, but a hole in the middle?
Any alteration of the image could be classified as distortion, but in this case the "distortions" can actually serve a purpose.
Mise
isle of lucy
I'm assuming (since you didn't specify) the object being imaged, the beams of light and the disks are pretty big, in which case (a) would have the image with a big black circle in the middle, and (b) would only have the middle part of the image.
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GoodGame
Red, White, & Blue, baby!
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Well, first off light can bend along an object (seen from holding a pencil up to a lamp). Probably the disk will absorb some of light that hits it, scatter some more, but some light will bend along the filter edges and still hit the image plane.
I think the in the first case the center of the image will be darker and more obscure relative to the outer edge of the image.
In the second case, I believe the light will just be more focused throught the filter, and the image will be unaffected, except that it will be much smaller than it was before the disk was placed, due to a smaller diameter of light reaching the image plane.
I'm assuming (since you didn't specify) the object being imaged, the beams of light and the disks are pretty big, in which case (a) would have the image with a big black circle in the middle, and (b) would only have the middle part of the image.
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No, that would be what happens if the disk is located in the object plane. If the disk is in the collimated beam, something different happens.
And yes, object, beam and disks can be assumed to be pretty big.
GoodGame said:
All true, but except for the extreme cases (where only a very small amount of light is not blocked) scattering and bending around effects can be neglected. Also the light source can be assumed to be incoherent, so interference doesn't play a role.
GoodGame said:
Depending on the object (a disk for example) this can be true. However, In the general case it's not.
No, the focus is not altered, the image will be affected and the size does not change.
No. The lenses are considered to be ideal lenses, they do nothing but collimate and focus the light.
I'll give a hint: There is an analogy to electronic signal processing, where the function of the disks can be realized by a capacitor and a resistor.
Mise
isle of lucy
It polarises the light? I'm guessing (a) and (b) polarise the light along different axes?
It filters the colours?
It filters the colours?
Perfection
The Great Head.
draw us a picture! 
Mise
isle of lucy
If it doesn't filter frequencies, does it filter amplitudes? So (a) will filter out everything brighter than something, and (b) will filter out everything dimmer than something?
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