SNELL'S LAW的問題!!!!!!!!!!!急!!!!!!!!!!!

2007-06-20 4:35 pm
我想問,點解speed of light in air/speed of light in glass=sin角of air/sin角of glass?!
有無人可以解釋比我知牙>..<謝謝^^

回答 (2)

2007-06-20 5:34 pm
✔ 最佳答案
這個問題我們需用到 Huygen's principle (屬於 A Level 課程) 來 prove:
Huygen's principle: Any point on a wavefront can be regarded as a secondary source of wavelet which progresses in the wave propagation direction at the same speed of the wave.
然後請見下圖:

圖片參考:http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Snell.jpg
參考: My physics knowledge
2007-06-20 5:04 pm
Since I cannot draw graph here, you need to have a graph of the light path first and to see my expanation.

Assume the light use time t to travel to the surface of the medium A and travel in the medium A for also time t, with incident angle (thetra air) and (thetra A)

distance travel in air (d Air) = ct, c is the speed of light.
distance travel in medium A (d A) = vA t, vA is the speed of light in the medium A.

In the light path, we can see the two triangle, the (d air) = L sin (thetra Air), where L is the centre line perpendicular the the surface of the mudium's boundary.
And (d A) = L sin (thetre A)

Then we have (d Air)/(d A)= ct/vA t = L sin (thetra Air)/ L sin (thetra A)
c/vA=sin (thetra Air)/sin (thetra A)

In fact, we have defined refractive index n, where n=c/vA,
n=sin (thetra Air)/sin (thetra A) => sin (thetra Air)=n sin (thetra A)

In gerneral, nB sin (thetra B)=nA sin (thetra A), for the light travel from medium B to A or A to B.


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