An optic fiber of refractive index 1.68 is coated with a protective covering of

Question

An optic fiber of refractive index 1.68 is coated with a protective covering of glass of refractive index 1.54.

Part A

Determine the critical angle for the fiber-glass interface

Part B

Determine the critical angle for the glass-air interface.

Part C

Determine the critical angle for a fiber-air interface (no glass covering).

Part D

Suppose a ray hits the fiber-glass interface at the angle calculated in part C. What is the angle of refraction of that ray when it reaches the glass-air interface?

Part E

Will the ray from part D leave the optic fiber?

Explanation / Answer

part A : the critical angle for fiber -glass inter face is

qc =sin-1(n2/n1)

here qc is critical angle , n2 is the refractive index of protective covering =1.54

n1 is the refractive index of fiber =1.68

qc =sin-1(n2/n1) ==sin-1(1.54/1.68) =66.44o

part B :critical angle for glass air inter face

   qc =sin-1(n(air)/n(glass)) = sin-1(1/1.54) =40.50o

Part C: critical angle for fiber air interface is

   qc =sin-1(n(air)/n(fiber)) = qc =sin-1(1/1.68) =36.53o

part D : by applying snell's law to fiber- glass interface

nfiber sin(i) =nglasssin(r)

here  nfiber =1.68 ,nglass =1.54 and i=36.53o

1.68 sin(36.53o) =1.54 sin(r)

r= sin-1(0.649) =40.49o

now applying snell's law for glass-air interface

nglass sin(i) = nair sin (r)

here ,nglass =1.54 ,  nair =1.00 and i =angle of incident =40.49o (from fiber-glass interface)

1.54 sin(40.49o ) =1.00 sin(r)

r=89.40o

angle of refraction when the ray reaches glass-air interface is 89.40o

part E :the critical angle for the glass-air interface is 40.50o (from part B) and the angle of incident of the ray in

glass-air interface is 40.49o this angle is almost equal to the critical angle so the ray will not escape from the optical fiber and it will travell parallel to the walls of optical fiber.

so answer is NO.

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