refractive index(40分)

Question

http://img81.imageshack.us/img81/6189/92079815hl8.jpg
ans:E

refractive index愈大即係spped愈細啦
光線由I去II係bends towards the normal
同埋光線由II去III有total internal reflection發生
所以II就係最optically denser than I同III,所以speed最細

但係我想問吓點樣比較I同埋III的refractive index?
有total internal reflection的發生都唔代表III係optically less dense than I果和...
ans點解係E唔可以係A?

Answer 1

我估差少少你便應該想到了! 既然 2 出唔到 3, 但你估 2 出唔出到 1?

當然是出到吧, 因為由 1 入 2, 然後反射回去的角度是一樣的, 所以 1 同 2 交界的兩條光線 同個 boundary 形成一個邊腰三角形, 即係話, 2 的光線折射出 1 那兒會係去返原本的角度出 1. 而因為 alternate angles, (boundary 1/2 // boundary 2/3), 由 normal 至 2/3 boundary 的反射角 與 normal 至 1/2 boundary 的入射角係一樣的! 所以同樣角度, 但出到 1, 出不到 3, 所以 3 比 1 更 optically less dense.

Answer 2

answer ofcourse is E.

we know that refactive index n increase will reduce the speed of light travelling in the medium, so I travel II bends. so n1 smaller than n2. Then II to III occurs total internal reflection, so n2 greater than n3. Then n1 or n3 is greater ?

we think the refracted ray as an incident ray which travel into III.
by alternate angle, the refracted angle of I to II = incident angle of II to III.

if n1smaller than n3, then no total internal reflection, and refracted angle of II to III is smaller than incident angle of I to II.
if n1 greater than n3, then critical angle of n3 is smaller than n1, so easy to occur total internal reflection.

We can conclude that n2>n1>n3, so speed is III>I>II