Use the following figure below for the next problem. The figure shows a cross- -A fiber optic is made by cladding a thin fiber core of refractive index n₁ = 1.45 with a material of refractive index n₂ = 1.38.What is the maximum incident angle,$\theta$,so that the light ray is totally internally reflected inside the fiber? A)24.1º B)26.4º C)28.2º D)30.3º E)32.7º

Question

Quizplus · Accepted Answer

Using Snell's law,
nsinθ = n1sinθ1
where n = refractive index of cladding, n1 = refractive index of core, θ = incident angle, and θ1 = angle of refraction (which will be 90° in this case for total internal reflection).
Rearranging the equation gives:
sinθ = (n1/n)sinθ1
θ = sin^(-1)((n1/n)sinθ1)
θ1 is the critical angle, which is the maximum incident angle for total internal reflection:
sinθ1 = n/n1
Substituting the values given,
sinθ1 = 1.45/1.38
sinθ1 = 1.0507
θ1 = sin^(-1)(1.0507)
θ1 = 82.18°
Substituting θ1 into the equation for θ,
θ = sin^(-1)((1.45/1.38)sin82.18°)
θ = 26.40° (rounded to two decimal places)
Therefore, the maximum incident angle for total internal reflection is 26.40°, which corresponds to choice B.

Use the Following Figure Below for the Next Problem θ\thetaθ ,So That the Light Ray Is Totally Internally Reflected Inside

Use the Following Figure Below for the Next Problem $\theta$ ,So That the Light Ray Is Totally Internally Reflected Inside