Ok have a look at the diagram(adsbygoogle = window.adsbygoogle || []).push({});

The diagram shows the end of an optical fibre. Inside is the core of n = 1.52 and the outer covering is the caldding n=1.45. For sufficiently small angles teh incident rays will be guided by the fiber and remain in teh core due to internal reflection. What is the maximum angle of incidence (theta) for which the light transmitted across the boundary will remain in the core??

I cn find the critical angle for which teh light will stay in teh core (i.e. the boundary of the core and the cladding) and that angle is 72.5 degres

so the angle the light inside the fiber must not exceed 17.5 degrees w.r.t. the base of the lining of the cladding.

So drawing some parallel lines the transmitted angle (refracted angle) into the core must not xceed 17.5 degrees

thus [tex] n_{1} sin \theta_{1} = n_{2} sin \theta_{2} [/tex]

since it is being transmitted from air to the core as the diagram shows

1 sin theta = 1.52 sin 17.5

theta = 27 degrees. So the angle of incidence must not exceed 17.5 degrees? Is this good ??

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# Homework Help: Help with optics

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