Tricky Questions!!!

Georgepowell

Joe has been showing Penny a few optical experiments. In one experiment he placed six mirrors vertically to form a regular hexagon with small gaps between the mirrors. Through one gap he shone a laser beam so it emerged straight from the gap diametrically opposite.

Penny then had to work out the smallest angle through which the beam must be rotated so that it emerged from the same gap as before, after being reflected just once by all six mirrors.

What was that small angle (rounded to the nearest degree)?

*DaveC426913*

Quote by Georgepowell

Joe has been showing Penny a few optical experiments. In one experiment he placed six mirrors vertically to form a regular hexagon with small gaps between the mirrors. Through one gap he shone a laser beam so it emerged straight from the gap diametrically opposite.

Penny then had to work out the smallest angle through which the beam must be rotated so that it emerged from the same gap as before, after being reflected just once by all six mirrors.

What was that small angle (rounded to the nearest degree)?

Upon what axis is the beam rotated? I presume the rotation axis can't be in line with the centre of the hexagon (i.e. spin the mirror contraption rather than the laser), so the axis of rotation must be the point where the laser beam first enters the hexagon?

Georgepowell

Quote by DaveC426913

Upon what axis is the beam rotated? I presume the rotation axis can't be in line with the centre of the hexagon (i.e. spin the mirror contraption rather than the laser), so the axis of rotation must be the point where the laser beam first enters the hexagon?

It is rotated about the point where it first enters the hexagon.

*DaveC426913*

I don't do answers. But I like diagramming questions!

Georgepowell

So is my question too difficult for the genius's of PhysicsForums.com ???

Rogerio

Quote by Georgepowell

Joe has been showing Penny a few optical experiments. In one experiment he placed six mirrors vertically to form a regular hexagon with small gaps between the mirrors. Through one gap he shone a laser beam so it emerged straight from the gap diametrically opposite.

Penny then had to work out the smallest angle through which the beam must be rotated so that it emerged from the same gap as before, after being reflected just once by all six mirrors.

What was that small angle (rounded to the nearest degree)?

Spoiler

9 degrees.
In fact, atan(sqrt(3)/11).

roam

It is given that there are two sets of real numbers A={a₁, a₂,..., a₁₀₀} and B={b₁, b₂,..., b₅₀}. If there is a mapping f from A to B such that every element in B has an inverse image and

f(a₁) ≤ f(a₂) ≤...≤ f(a₁₀₀)

Then the number of such mapping is...

A) C⁵⁰₁₀₀

B) C⁵⁰₉₉

C) C⁴⁹₅₀

D) C⁴⁹₉₉

Georgepowell

Quote by Rogerio

Spoiler

9 degrees.
In fact, atan(sqrt(3)/11).

why? what was your logic?

Rogerio

Quote by Georgepowell

why? what was your logic?

I hope it is enough.

*DaveC426913*

Quote by Rogerio

I hope it is enough.

Wow. What a brilliant way of solving it.

redargon

I don't get it?

roam

The answer is (D): C⁴⁹₉₉

cool_arrow

Hi everyone. I saw this problem in a logic textbook I had in a class many years ago:

Two friends meet after having been out of contact with one another for some
years. They have the the following conversation:

A: I have three sons.
B: What are their ages?
A: The product of their ages is 36.
B: That is not enough information.
A: The sum of their ages is the same as the the number on the building across the street.
B: Give me a minute to work it out with pencil and paper.

B: I've almost got it but I need one more clue.
A: The oldest one has red hair.
B: I've got it.

What are the ages of the three sons?

Rogerio

Quote by cool_arrow

Hi everyone. I saw this problem in a logic textbook I had in a class many years ago:

Two friends meet after having been out of contact with one another for some
years. They have the the following conversation:

A: I have three sons.
B: What are their ages?
A: The product of their ages is 36.
B: That is not enough information.
A: The sum of their ages is the same as the the number on the building across the street.
B: Give me a minute to work it out with pencil and paper.

B: I've almost got it but I need one more clue.
A: The oldest one has red hair.
B: I've got it.

What are the ages of the three sons?

Well, it is a classic puzzle, and I've seen it a couple of times, right in this forum...

http://www.physicsforums.com/showthread.php?t=50813

Rogerio

Quote by DaveC426913

Wow....

Thanks, Dave!

redargon

Quote by Rogerio

Thanks, Dave!

I still don't get it, could you explain your diagram a little please?

Rogerio

Quote by redargon

I still don't get it, could you explain your diagram a little please?

Just try to follow the ray, using geometry.

The figure shows the first reflection.
Note the first mirror surface (blue line) , its normal (yellow line) , the reflected ray (pink line) , and the simmetrical points (cyan).

After the first reflection you have to follow the pink line. However, everything goes as you were at the second hexagon following the red line, and so on.

Just follow the ray, and you'll understand!

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DaveC426913	I don't do answers. But I like diagramming questions! Attached Thumbnails

Georgepowell	So is my question too difficult for the genius's of PhysicsForums.com ???

roam	It is given that there are two sets of real numbers A={a₁, a₂,..., a₁₀₀} and B={b₁, b₂,..., b₅₀}. If there is a mapping f from A to B such that every element in B has an inverse image and f(a₁) ≤ f(a₂) ≤...≤ f(a₁₀₀) Then the number of such mapping is... A) C⁵⁰₁₀₀ B) C⁵⁰₉₉ C) C⁴⁹₅₀ D) C⁴⁹₉₉