Question

The indices of refraction of the core and cladding of an optical fiber are 1.5 and 1.40, respectively. Light is incident at the core at 70%. What is the distance between two successive reflections in the core it its diameter is 0.05mm?

Answer 1

As per the question,

Refractive index of core $(\mu_1)=1.5$

Refractive index of cladding $(\mu_2)=1.4$

Angle of incidence $= 70^\circ$

Diameter of the core$=0.05mm$

$\Rightarrow L=d\sqrt{(\dfrac{\mu_2}{\mu_1\sin\theta})^2-1}$

$\Rightarrow L=0.05\sqrt{(\dfrac{1.5}{1.4\sin(70)})^2-1}$

$\Rightarrow L =0.05\sqrt{1.27-1}=0.05\sqrt{0.27}=0.025mm$