**10 figure h shows a straight optical fibre. the reflective indices of the core and the cladding are ( n_1…

**10 figure h shows a straight optical fibre. the reflective indices of the core and the cladding are ( n_1 ) and ( n_2 ) respectively, and ( n_1>n_2 ).\nfor a ray of light passing from one end of the fibre to the other end without leaking from the core, if the minimum time needed is ( t ), what is the maximum time needed?\na ( \frac{t}{n_1} )\nb ( \frac{t}{n_2} )\nc ( \frac{n_1}{n_2}t )\nd ( \frac{n_2}{n_1}t )

**10 figure h shows a straight optical fibre. the reflective indices of the core and the cladding are ( n_1 ) and ( n_2 ) respectively, and ( n_1>n_2 ).\nfor a ray of light passing from one end of the fibre to the other end without leaking from the core, if the minimum time needed is ( t ), what is the maximum time needed?\na ( \frac{t}{n_1} )\nb ( \frac{t}{n_2} )\nc ( \frac{n_1}{n_2}t )\nd ( \frac{n_2}{n_1}t )

Answer

Explanation:

Step1: 求光在光纖中的速度

根據公式(v=\frac{c}{n})（(c)為真空中光速，(n)為折射率），光在核心層速度(v_1 = \frac{c}{n_1})。 當光沿光纖軸線直傳（無反射）時，路程(L)最短，時間(T=\frac{L}{v_1}=\frac{n_1L}{c})。

Step2: 求臨界角(\theta_c)

由全反射臨界角公式(\sin\theta_c=\frac{n_2}{n_1})。 設光纖長度為(L)，當光以臨界角不斷反射傳播時，光線路程(s=\frac{L}{\sin\theta_c})，將(\sin\theta_c=\frac{n_2}{n_1})代入得(s = \frac{n_1L}{n_2})。

Step3: 求最長時間

光在核心層速度(v_1=\frac{c}{n_1})，根據(t=\frac{s}{v_1})，把(s = \frac{n_1L}{n_2})和(v_1=\frac{c}{n_1})代入，(t=\frac{\frac{n_1L}{n_2}}{\frac{c}{n_1}}=\frac{n_1^2L}{n_2c})。 又因為(T=\frac{n_1L}{c})，所以(t = \frac{n_1}{n_2}T)。

Answer:

C. (\frac{n_1}{n_2}T)