Answer to Question #116610 in Electricity and Magnetism for Noemy

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$=0.05mm$

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

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

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