How Does Light Behave at the Critical Angle in an Optical Fiber?

[Nb]

The drawing shows an optical fiber that consists of a core made of flint glass (nflint = 1.667) surrounded by a cladding made of crown glass (ncrown = 1.523). A beam of light enters the fiber from air at an angle q 1 with respect to the normal. What is q 1 if the light strikes the core-cladding interface at the critical angle q c?http://edugen.wiley.com/edugen/courses/crs1000/art/images/c26/nw1183-n.gif

i know u have to use snell's law

sin theta = (n2/n1), but i don't know what to do from here.

The textbook in which this problem is in gives no info on these kinds problems. Please help me

 

[quantumdude]

You have to use Snell's law twice. You have to do it once to find the critical angle (considering the cladding-core interface) and then you have to do it again to find θ₁ (considering the core-air interface).

 

[M98Ranger]

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Total internal reflection occurs when a light ray strikes a medium boundary at an angle greater than the critical angle, causing the light to be completely reflected back into the original medium. In this case, the medium boundary is the core-cladding interface of the optical fiber.

To find the critical angle, we can use Snell's law, which states that the ratio of the sine of the angle of incidence (q1) to the sine of the angle of refraction (q2) is equal to the ratio of the refractive indices of the two media.

So we have: sin q1 / sin q2 = n2 / n1

Since the light is entering from air (n1 = 1), we can rewrite this as: sin q1 / sin q2 = n2 / 1

Now, at the critical angle, the angle of refraction is 90 degrees, so sin q2 = 1. Therefore, we can rewrite the equation as: sin q1 = n2

Substituting in the refractive indices for flint glass (nflint = 1.667) and crown glass (ncrown = 1.523), we get: sin q1 = 1.667

To solve for q1, we can take the inverse sine of both sides: q1 = sin^-1 (1.667) = 69.5 degrees.

So, if the light strikes the core-cladding interface at an angle of 69.5 degrees, it will be at the critical angle and will undergo total internal reflection. Any angle greater than this will also result in total internal reflection.