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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?
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Verified Answer
The correct answer is:
$0^{\circ}-30^{\circ}$
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.
$\sin \theta=\frac{\sqrt{\mu_1^2-\mu_2^2}}{\mu_0}$
$\theta=\sin ^{-1} \sqrt{\left(\mu_1\right)^2-\left(\mu_2\right)^2} \quad\left[\right.$ 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.
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