Q133 / 160 · AP EAMCET 2010 · Physics

In an optical fibre, core and cladding were made with materials of refractive indices 1.5 and 1.414 respectively. To observe total internal reflection, what will be the range of incident angle with the axis of optical fibre?

  • A
    $0^{\circ}-60^{\circ}$
  • B
    $0-48^{\circ}$
  • Correct answer
    C
    $0^{\circ}-30^{\circ}$
  • D
    $0^{\circ}-82^{\circ}$

Solution

The range of incident angle $\sin \theta=\frac{\sqrt{\mu_1^2-\mu_2^2}}{\mu_0}$ $\theta=\sin ^{-1} \sqrt{\left(\mu_1\right)^2-\left(\mu_2\right)^2} \quad\left[\right.$ for air $\left.\mu_0=1\right]$ $\theta=\sin ^{-1} \sqrt{(1.5)^2-(1.414)^2}$ $=\sin ^{-1}(0.5006)$ $\theta=30^{\circ}$ Hence, option (c) is correct.

Hence, the correct answer is option: 'C'.

Relevant Questions