Physicists' Laser Show
Consider yourself strolling through a unique and exciting carnival.
Suddenly you come across an interesting amusement labeled 'Laser Target Shoot' . The attendant guides you to a small rectangular chamber with walls covered with ideally reflecting mirrors. At your corner ( Corner 1 ), there is a powerful laser locked into place, horizontally at an angle 45 to the walls. The attendant asks you to fire a shot.
The length of the chamber is 3536 cm while its breadth is 697 cm. Targets are placed at all the 4 corners of the room (including CORNER 1 , from where the laser was fired)
Can you find out the number of reflections the laser beam will undergo before finally hitting one of the four targets and which target will it finally hit?
If the total number of reflections is and the corner number of the target which the laser will finally hit is , give your answer as .
Details and Assumptions
- All the 4 surfaces (walls) are perfectly reflecting.
- There is no loss of intensity of the laser beam whatever be the number of reflections.
- Targets have been placed at all the four corners i.e 1,2, 3 and 4 as marked in the figure.
- The laser beam will be fired only at the given angle and the given position i.e. it cannot be adjusted.
P.S. This is simple enough to be done manually, I don't approve of a computer-based solution.
The answer is 251.
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1 solution
The problem requires not a bit of knowledege of Number Theory, I guess , only a pinch of creativity.
Imagine...... For a moment, let us consider the path that the light will travel across the chamber as a the path travelled by light in a continous chamber of any length, which we will decide later.
Now, if we join two chambers as shown such that the common wall (wall 3-4) is removed such that the light is free to pass through, then we can see that after the light ray strikes wall 3-4, the path it would have travelled next till it strikes wall 1-2 (represented by darkened continous pencil) in chamber A in the original (unjoined) chamber, it will be same as the light travelling in chamber B (represented by dotted pencil) after the wall is removed. As we can see the path 1 (the darkened continuous pencil) in chamber A, will be the mirror image symmetrical image of path 2 (dotted pencil), so we can mark the other corners of chamber B as 2 and 1 as I did.
So we can trace out the path that the light will travel in the following way.
Clearly we can see that the the horizontal distance that the light will cover in two consecutive reflections is 6 9 7 as tan 4 5 = 1 as I have represented. So, obviously, we need to find number of chambers that we need to join to make light strike a corner which clearly can be done by finding the least common multiple of 6 9 7 and 3 5 3 6 which comes out to be 1 4 4 9 7 6 , so we need 41 chambers. As with each reflection the ray covers 6 9 7 so there are in total 6 9 7 1 4 4 9 7 6 − 1 = 2 0 7 reflections. We substracted 1 because the last reflection when the light hits the corner shall not be counted. But we still need to add the reflections in vertical walls which one can eisily figure out is 4 0 as there will be 4 0 common walls between the chambers through which the light will pass which gives us the answer, 2 0 7 + 4 0 = 2 4 7
Now we need to find out the corner in whch the ray will strike. As the ray will stikes the bottom horizintal wall , wall 1-4 after covering every horizintal distance of 1 3 9 4 and as 1 3 9 4 ∣ 1 4 4 7 9 6 so clearly the ray will strike one of the bottom corners 1 or 4. Now, the right wall of the 4 1 r s t chamber will be 3-4 so the corner which the ray will strike is 4.
Answer to the question, 2 4 7 + 4 = 2 5 1
B I N G O ! ! !