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Find angle required for total internal reflection of a beam of light

  • Thread starter MrMoose
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  • #1
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Homework Statement



An optical fiber consists of a glass core (index of refraction n1) surrounded by a coating (index of refration n2<n1). Suppose a beam of light enters the fiber from air at an angle θ with the fiber axis as shown in attached. (a) Show that the greatest possible value of θ for which a ray can travel down the fiber is given by θ = arcsin(sqrt(n1^2 - n2^2)). (b) If the indices of refraction of the glass and coating are 1.58 and 1.53, respectively, what is this value of the incident angle θ?

Homework Equations



na*sin(θa) = n1*sin(θ1)

Critical Angle: θc = arcsin(n1/n2)

The Attempt at a Solution



See attached for my drawing. Any angle greater than θc will cause total internal reflection so,

θo > θc

where θo = 90 - θ1

Substituting for θc

θo > arcsin(n2/n1)

Substituting for θo

90 - θ1 > arcsin(n2/n1)

Substituting for θ1

90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)

Taking sine of the whole equation

1 - sin(θa/n1) > n2/n1

Solving for θa

θa < arcsin(n1-n2)

According to my calculations, this would be the maximum angle θa for which a ray can travel down the fiber. Unfortunately it's not correct.

Where do they get the sqrt(n1^2-n2^2) term?

Thanks in advance, MrMoose
 

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Answers and Replies

  • #2
ehild
Homework Helper
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1,803

Homework Statement



An optical fiber consists of a glass core (index of refraction n1) surrounded by a coating (index of refration n2<n1). Suppose a beam of light enters the fiber from air at an angle θ with the fiber axis as shown in attached. (a) Show that the greatest possible value of θ for which a ray can travel down the fiber is given by θ = arcsin(sqrt(n1^2 - n2^2)). (b) If the indices of refraction of the glass and coating are 1.58 and 1.53, respectively, what is this value of the incident angle θ?

Homework Equations



na*sin(θa) = n1*sin(θ1)

Critical Angle: θc = arcsin(n1/n2)

The Attempt at a Solution



See attached for my drawing. Any angle greater than θc will cause total internal reflection so,

θo > θc

where θo = 90 - θ1

Substituting for θc

θo > arcsin(n2/n1)

Substituting for θo

90 - θ1 > arcsin(n2/n1)

Substituting for θ1

90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)

Taking sine of the whole equation

1 - sin(θa/n1) > n2/n1


Solving for θa

θa < arcsin(n1-n2)

According to my calculations, this would be the maximum angle θa for which a ray can travel down the fiber. Unfortunately it's not correct.

Where do they get the sqrt(n1^2-n2^2) term?

Thanks in advance, MrMoose
The equation in red is wrong. sin(90-ψ) is not 1-sinψ.

ehild
 
  • #3
23
0
Nice, Thanks ehild! I think I got it now. Taking up from just before the incorrect equation:

90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)

sin[90 - arcsin(sin(θa)/n1)] > n2/n1

using the following property for sine: sin(x-y) = sinx*cosy - cosx*siny

sin(90)*cos[arcsin(sinθa/n1)] - sin(θa)/n1*cos(90) > n2/n1

Simplifying:

cos[arcsin(sinθa/n1)] > n2/n1

using the following property for cosine: cos[arcsin(x)] = sqrt(1-x^2)

sqrt{1-[sin(θa)/n1]^2} > n2/n1

solving for θa.

θa < arcsin[sqrt(n1^2-n2^2)]

Thanks for your help!
 
  • #4
ehild
Homework Helper
15,396
1,803
Good job!

sin[90 - arcsin(sin(θa)/n1)] > n2/n1

using the following property for sine: sin(x-y) = sinx*cosy - cosx*siny
It is correct. Even simpler: the cosine of an angle θ in a right triangle is just the sine of the other angle, 90-θ. :smile:

ehild
 
  • #5
433
7

Homework Statement



An optical fiber consists of a glass core (index of refraction n1) surrounded by a coating (index of refration n2<n1). Suppose a beam of light enters the fiber from air at an angle θ with the fiber axis as shown in attached. (a) Show that the greatest possible value of θ for which a ray can travel down the fiber is given by θ = arcsin(sqrt(n1^2 - n2^2)). (b) If the indices of refraction of the glass and coating are 1.58 and 1.53, respectively, what is this value of the incident angle θ?

Homework Equations



na*sin(θa) = n1*sin(θ1)

Critical Angle: θc = arcsin(n1/n2)

The Attempt at a Solution



See attached for my drawing. Any angle greater than θc will cause total internal reflection so,

θo > θc

where θo = 90 - θ1

Substituting for θc

θo > arcsin(n2/n1)

Substituting for θo

90 - θ1 > arcsin(n2/n1)

Substituting for θ1

90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)


Taking sine of the whole equation

1 - sin(θa/n1) > n2/n1

Solving for θa

θa < arcsin(n1-n2)

According to my calculations, this would be the maximum angle θa for which a ray can travel down the fiber. Unfortunately it's not correct.

Where do they get the sqrt(n1^2-n2^2) term?

Thanks in advance, MrMoose
It may be painfully obvious, but how exactly was θ1 = arcsin(sin(θa)/n1) derived?
 

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