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?

Question

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°-60° (B) 0-48° (C) 0°-30° (D) 0°-82°

Tardigrade Admin · Accepted Answer

The range of incident angle sin θ =(√μ12-μ22/μ0) θ= sin -1 √(μ1)2-(μ2)2 [for air μ0=1] θ= sin -1 √(1.5)2-(1.414)2 = sin -1(0.5006) θ =30°

Q. 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?

$0^{\circ} - 6 0^{\circ}$

$0 - 4 8^{\circ}$

$0^{\circ} - 3 0^{\circ}$

$0^{\circ} - 8 2^{\circ}$

The range of incident angle
$sin θ = \frac{μ _{1}^{2} - μ _{2}^{2}}{μ _{0}}$
$θ = sin^{- 1} (μ_{1})^{2} - (μ_{2})^{2}$ [for air $μ_{0} = 1$ ]
$θ = sin^{- 1} (1.5)^{2} - (1.414)^{2}$
$= sin^{- 1} (0.5006)$
$θ = 3 0^{\circ}$

Q. 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?

0∘−60∘

0−48∘

0∘−30∘

0∘−82∘