Question

An optical fibre has a cylindrical cross-section of diameter $$ d $$ and index of refraction $$ n $$ bent sharply. What is the smallest radius of curvature at a short bent section for which the total internal reflection will be assured for light initially travelling parallel to the axis of the fibre?

Answer 1

Correct option is A. $$\dfrac{nd}{n-1}$$
Total internal reflection takes place for all angles greater than a critical angle. Thus, for the smallest radius of curvature, the angle made at the right end of the fibre $$(i)$$ should be equal to the critical angle, for total internal reflection. Thus,

$$\dfrac{\sin90^{\circ}}{\sin i} = \dfrac{n_1}{n_2}$$

But, $$n_2 = 1$$, and $$n_1 = n$$

$$\Rightarrow \sin i = \dfrac{1}{n}$$

$$\Rightarrow \dfrac{r-d}{r} = \dfrac{1}{n}$$

$$\Rightarrow n(r-d) = r$$

$$\Rightarrow nr - r = nd$$

$$\Rightarrow r(n-1) = nd$$

$$\therefore r = \dfrac{nd}{n-1}$$