i have started something, dont know if its right.
I know that
sin(θ1) = L1/CX
CX = L1/sinθ
sin(θ2) = L2/XD
XD = L2/sin(θ2)
CXD = (L1/sin(θ1)) + (L2/sin(θ2))
hmm and now im stuck again
I’ve been working on this task for a while but i don’t know where to start.
Anyone that can give me any tips on how to solve it?
Light travels in such a way that it requires the minimum possible time to get from one point to another. A ray of light from C reflects of a plane mirror AB at X and then passes through D. (see figure) Show that the rays CX and XD make equal angles with the normal to AB at X. (Remark: you may wish to give a proof based on elementary geometry without using any calculus, or you can minimize the travel time on CXD.)
Warning: I'm just learning geometry, so some of this may be wrong (that's for someone else to decide)...
Make a point on line AB so that it is directly under point C, label it E
Now continue ray CX past line AB
Now place a point on ray CX so that it is directly under point D, label it F
Make a point on line AB so that it is directly above point F, label it G
Now angles AXC and BXF are vertical, thus congruent.
This means that triangles CXE and FXG are similar
Now when the line hits the mirror, it's path is reflected across the line AB
Thus, reflect triangle FXG across the line (you'll get triangle DGX)
Triangles DGX and CXE are still similar
Thus angles BXD and AXC are congruent.
Thank you for a fast answer.
Your solution seems to be correct.
Now I only need to formulate an answer, to show that the relation between the triangels proves that the shortest way between CXD is when angles BXD and AXC are congruent.
Shouldn’t bee to hard though if I assume that the fastest way between two points is the straight way. This is the case if the hypotenuse of triangel FXG and GXD is the same.
If anyone has something further to add I’m glad to hear.
The path taken is a stationary point of the time for a photon to travel
from the source to the detector. The direct path and the reflected path
are both stationary in the sense that the travel time along these paths
is less than that along any other path close to them.
Of course the direct path in this case takes less time and is shorter than
the reflected path. But they are both local minima for travel time.
RonL
Hi,
if you only have the points C and D and the plane AB then you find the point X by reflecting D at AB to D'. Connect the points C and D'. The intersection of CD' and AB gives point X. (I've attached your modified diagram to show what I mean).
You now are dealing with isoscele triangles so you can easily proof the values of the angles.
EB