Distance Optimisation and Optics
Maths in english
Split in groups of 3 or 4 to tackle these optimisation problems. During your research, talk only in english.
John wants to ride his horse from point A to point B shown on the map below (one unit represents one mile) but he will have to stop by the river (colored in grey) to let his horse drink. He wonders where he should stop to make his trip as short as possible. Can you help him? Problem 1:
4
b
B
3 2
A
b
1 b
b
1
2
3
4
5
6
7
8
9
−1 b
−2 b
David Hasselhoff (point H on the map) is a lifeguard on Malibu beach. One day, while staring at the sea (in grey), he saw someone drowning at point B. He had to get there as fast as he could to rescue him! But David H. has never been so good at maths, the only thing he could think of is that the shortest route between two points is the straight line. So he went straight forward to point B. But there is one fact he didn’t take into account: Problem 2:
He runs faster than he swims! Assuming that it takes him one second to run one unit and that he runs twice as fast as he swims, show that he did not follow the best route. 2 bH 1 b
b
1
2
3
4
5
6
7
−1 −2
b
B
b
http://prof.pantaloni.free.fr
Lyc´ee Jean Zay, Orl´eans b
Maths in english
Distance Optimisation and Optics
Back on problem 1 As a homework for today you had to explain the trick using a symetry to find point C where John had to stop his horse. Now draw a line L passing through C and perpendicular to the river. Let N be a point on L whose y-value is positive. \ and N \ 1. What can you say about these two angles: ACN CB? (Prove it!) 2. Could you find a physical situation where this also occurs? (There are at least two possible examples) 3. Do you think this is sheer luck? Back on problem 2 Light has several properties, we will consider these three: ① In a any homogeneous medium, light travels in straight lines. That’s why we can modelize it’s route by rays of light. ② Fermat’s principle: In optics, Fermat’s principle or the principle of least time is the principle that the path taken between two points by a ray of light is the path that can be traversed in the least time. ③ Light’s speed is different depending on the transparent medium it goes through. For instance, in air the speed of light is 3 × 108 m/s whereas in water it is ”only” 2.25 × 108 m/s Now split in groups of 3 or 4 and work on one of these tasks. Then you’ll change groups. 1. (3 people) Using GeoGebra, draw a figure for problem 2 and try to find the best possible point where David Hasseloff should dive. 2. Calculate the time function T (x) giving the time of David’s travel depending on where he dives. Plot it on your calculator and try to find an answer to the problem using: a. A table of values, b. A graphical solver (Calc/minimum for TI or Gsolve/minimum for Casio) 3. Find an optical analogy of problem 2. Try to write it using the words: ray, light speed, Fermat’s principle, surface of water, direction, angle. . . Draw a sketch with rays and angles, similar to what we did for problem 1. Do you think we will have equal angles as well?
http://prof.pantaloni.free.fr
Lyc´ee Jean Zay, Orl´eans
Maths in english
Distance Optimisation and Optics
Video on You Tube: Refraction Part1 Try to fill in the dots with the missing words from the video you just saw. Ptolemy’s experience:
Nearly . . . . . . . . . . . . years ago, the roman philosopher and . . . . . . . . . . . . Ptolemy tried to explain how light . . . . . . . . . . . . as it passes from one. . . . . . . . . . . . to another. In a treatise on. . . . . . . . . . . . he described a . . . . . . . . . . . . of light which you can easily repeat today. Ptolemy put a . . . . . . . . . . . . on a table. Inside the bowl he placed a . . . . . . . . . . . . . Then Ptolemy . . . . . . . . . . . . his eye until the coin was no longer . . . . . . . . . . . . over the. . . . . . . . . . . . of the bowl. The next step was to . . . . . . . . . . . . water into the bowl. As the water level. . . . . . . . . . . . in the bowl, the . . . . . . . . . . . . of the coin became visible. We call this refraction, the bending of light as it passes between two transparent substances. An . . . . . . . . . . . . ray from the object – the coin – . . . . . . . . . . . . the surface of the water. The . . . . . . . . . . . . ray changes direction to reach the eye. The image of the coin . . . . . . . . . . . . on the . . . . . . . . . . . . of the refracted ray . . . . . . . . . . . . . . . . . . . . . the liquid, even as the coin itself lies out of. . . . . . . . . . . . . . . . . . . . . . . . to the eye, below the . . . . . . . . . . . . of the bowl. Draw a sketch of Ptolemy’s experience with the light rays, explaining the trick.
: In which year did Snell finally discover a law on refraction? . . . . . . . . . What is a line normal to the surface of water? . ................................................................... Where do all these three lines lie: incident ray, refracted ray and normal to the surface? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . What is the ”sine of the angle θ ”? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The index of refraction is the ratio what? . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................... Snell’s law of refraction
http://prof.pantaloni.free.fr
Lyc´ee Jean Zay, Orl´eans
Distance Optimisation and Optics
Maths in english
In
normal
Nearly 2000 years ago, the roman philosopher and scientist Ptolemy tried to explain how light bends as it passes from one medium to another. In a treatise on optics he described a trick of light which you can easily repeat today. Ptolemy put a bowl on a table. Inside the bowl he placed a coin. Then Ptolemy lowered his eye until the coin was no longer visible over the rim of the bowl. The next step was to pour water into the bowl. As the water level rose in the bowl, the image of the coin became visible. We call this refraction, the bending of light as it passes between two transparent substances. An incident ray from the object – the coin – strikes the surface of the water. The refracted ray changes direction to reach the eye. The image of the coin lies on the extension of the refracted ray back into the liquid, even as the coin itself lies out of line of sight to the ye, below the lip of the bowl.
ci nt de
θ1
y ra interface
v1
O
v2 ref
Air
rac ted ray
θ2 Y
Water
In
normal
X
ci nt de
θ1
y ra interface
v1
O
v2 ref
Air
rac
X
ray
Y
ted
θ2
Water
http://prof.pantaloni.free.fr
Lyc´ee Jean Zay, Orl´eans
