Question

An optical fiber consists of an outer "cladding" layer and an inner core with a slightly higher index of refraction. Light rays entering the core are trapped inside by total internal reflection and forced to travel along the fiber (see the following figure). Suppose the cladding has an index of refraction of 1.46 and the core has an index of refraction of 1.48.Calculate the largest angle between a light ray and the longitudinal axis of the fiber for which the ray will be totally internally reflected at the core/cladding boundary.

Answer 1

Answer

given,

refractive index of cladding = 1.46

refractive index of inner core = 1.48

largest angle between light ray = ?

critical angle,

$\theta_c = sin^{-1}(\dfrac{1.46}{1.48})$

$\theta_c =80.6^0$

$\theta_c =90^0 - 80.6^0= 9.4^0$

According to Snell's law

    $n_1 sin\theta_i= n_2 sin \theta_r$

    $sin\theta_i = 1.48\times sin 9.4^0$

    $\theta_i = sin^{-1}{0.242}$

    $\theta_i =14^0$

assuming angle of incident be equal to 50°

then Largest angle

        =50° - 14° = 36°