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Q. Refer figure, core has refractive index $\mu_1 = 1.424$. The cladding refractive index $\mu_2 = 1.39$. In such a case, will the light beam propagate?Physics Question Image

Communication Systems

Solution:

If $\mu_1$ is the refractive index of core of optical fibre, then
$\mu_{1} = \frac{sin\,10^{\circ}}{sin\,7^{\circ }} = \frac{0.1736}{0.1219} = 1.424$
Critical angle at core and cladding surface is
$sin\,C = \frac{\mu _{2}}{\mu _{1}} = \frac{1.39}{1.424} = 0.9761 = sin\,77^{\circ } 27'$
$C = 77^{\circ } 27'$
The angle of incidence on core-cladding surface $= 90^{\circ} - 7^{\circ} = 83^{\circ}$, which is greater than critical angle, hence beam will propagate through optical fibre.