An optical fiber has an index of refraction n and diameter d . It is surrounded by vacuum. Light is sent into the fiber along its axis as shown in Figure P34.31. (a) Find the smallest outside radius R min permitted for a bend in the fiber if no light is to escape. (b) What If? What result does part (a) predict as d approaches zero? Is this behavior reasonable? Explain. (c) As n increases? (d) As n approaches 1? (c) Evaluate R min assuming the fiber diameter is 100 μ m and its index of refraction is 1.40. Figure P34.31

Question

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Answer 1

Textbook Question

Chapter 35, Problem 35.49P

An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is sent into the fiber along its axis as shown in Figure P34.31. (a) Find the smallest outside radius R_min permitted for a bend in the fiber if no light is to escape. (b) What If? What result does part (a) predict as d approaches zero? Is this behavior reasonable? Explain. (c) As n increases? (d) As n approaches 1? (c) Evaluate R_min assuming the fiber diameter is 100 μm and its index of refraction is 1.40.

Figure P34.31

$Chapter 35, Problem 35.49P, An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is$

(a)

Expert Solution

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

(b)

Expert Solution

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

(c)

Expert Solution

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

(d)

Expert Solution

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

(e)

Expert Solution

To determine

The value of

$Rmin$ .

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

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Answer 2

Textbook Question

Chapter 35, Problem 35.49P

An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is sent into the fiber along its axis as shown in Figure P34.31. (a) Find the smallest outside radius R_min permitted for a bend in the fiber if no light is to escape. (b) What If? What result does part (a) predict as d approaches zero? Is this behavior reasonable? Explain. (c) As n increases? (d) As n approaches 1? (c) Evaluate R_min assuming the fiber diameter is 100 μm and its index of refraction is 1.40.

Figure P34.31

$Chapter 35, Problem 35.49P, An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is$

(a)

Expert Solution

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

(b)

Expert Solution

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

(c)

Expert Solution

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

(d)

Expert Solution

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

(e)

Expert Solution

To determine

The value of

$Rmin$ .

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

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Answer 3

Textbook Question

Chapter 35, Problem 35.49P

An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is sent into the fiber along its axis as shown in Figure P34.31. (a) Find the smallest outside radius R_min permitted for a bend in the fiber if no light is to escape. (b) What If? What result does part (a) predict as d approaches zero? Is this behavior reasonable? Explain. (c) As n increases? (d) As n approaches 1? (c) Evaluate R_min assuming the fiber diameter is 100 μm and its index of refraction is 1.40.

Figure P34.31

$Chapter 35, Problem 35.49P, An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is$

(a)

Expert Solution

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

(b)

Expert Solution

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

(c)

Expert Solution

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

(d)

Expert Solution

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

(e)

Expert Solution

To determine

The value of

$Rmin$ .

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

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Answer 4

Textbook Question

Chapter 35, Problem 35.49P

An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is sent into the fiber along its axis as shown in Figure P34.31. (a) Find the smallest outside radius R_min permitted for a bend in the fiber if no light is to escape. (b) What If? What result does part (a) predict as d approaches zero? Is this behavior reasonable? Explain. (c) As n increases? (d) As n approaches 1? (c) Evaluate R_min assuming the fiber diameter is 100 μm and its index of refraction is 1.40.

Figure P34.31

$Chapter 35, Problem 35.49P, An optical fiber has an index of refraction n and diameter d. It is surrounded by vacuum. Light is$

(a)

Expert Solution

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

(b)

Expert Solution

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

(c)

Expert Solution

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

(d)

Expert Solution

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

(e)

Expert Solution

To determine

The value of

$Rmin$ .

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

(b)

Expert Solution

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

(c)

Expert Solution

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

(d)

Expert Solution

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

(e)

Expert Solution

To determine

The value of

$Rmin$ .

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

Answer 8

(a)

Expert Solution

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape.

Answer 13

Answer to Problem 35.49P

The smallest outside radius

$Rmin$ . permitted for a bend in the fiber if no light is to escape is

$ndn−1$ .

Answer 14

Explanation of Solution

Given info: The index of refraction is $n$ , the diameter of the optical fiber is $d$ , it is surrounded by vacuum.

The figure that represents the given conditions is shown below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.49P

Figure (1)

From figure a ray originally moves along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

The necessary condition for the ray to be total internal reflection is,

$θ0≥(θ0)c$

Here,

$θ0$ is the angle of incidence.

$(θ0)c$ is the critical angle.

For very small value of $θ0$ .

The above condition can be written as,

$sinθ0≥sin(θ0)c$ (1)

The expression of $sin(θ0)c$ is,

$sin(θ0)c=nairnpipe$

Substitute 1 for the value of $nair$ , and $n$ for the value of $npipe$ in the above expression.

$sin(θ0)c=1n$

The expression of $sinθ0$ is,

$sinθ0=R−dR$

Here,

$R−d$ is the distance of optical fiber from the center.

$R$ is the outside radius.

The expression of equation (1) becomes.

$sinθ0≥sin(θ0)cR−dR≥1nR≥ndn−1$ .

The minimum value of outside radius permitted for a bend in the fiber is $ndn−1$ .

Conclusion:

Therefore, the smallest outside radius $Rmin$ . permitted for a bend in the fiber if no light is to escape is $ndn−1$

Answer 15

(b)

Expert Solution

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The effect to result on part (a) as

$d$ approaches to zero and to analyze the behavior is reasonable or not.

Answer 20

Answer to Problem 35.49P

The

$Rmin→0$ . Yes, for very small

$d$ the light strikes the interface at very large angles of incidence.

Answer 21

Explanation of Solution

The result from part (a) is,

$R≥ndn−1$ (1)

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The value of $d$ becomes very small or goes to zero than $Rmin$ will tends to zero from equation (1).

The lesser value of $d$ indicates the optical fiber becomes thinner, if the optical fiber becomes thinner than the angle of incidence becomes very large and it strikes the surface at very large angle of incidence.

The thinner the optical fiber, the radius up to which the fiber is bent becomes smaller.

Conclusion:

Therefore, the $Rmin→0$ and yes, for very small $d$ light strikes the interface at very large angles of incidence.

Answer 22

(c)

Expert Solution

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The effect to result on part (a) as

$n$ increases and to analyze this behavior is reasonable or not.

Answer 27

Answer to Problem 35.49P

The

$Rmin$ decreases and yes, as

$n$ increases the critical angle becomes smaller.

Answer 28

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1=d1−1n$

It is clear that as $n$ increases $Rmin$ will decrease.

Yes, as $n$ increases the critical angle becomes smaller

Conclusion:

Therefore, the $Rmin$ decreases and yes, as $n$ increases the critical angle becomes smaller.

Answer 29

(d)

Expert Solution

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The effect to result on part (a) as

$n$ approaches to

$1$ and to analyze this behavior is reasonable or not.

Answer 34

Answer to Problem 35.49P

The

$Rmin→∞$ and yes, as

$n$ approaches to

$1$ , the critical angle becomes close to

$90°$ and any bend will allow the light to escape.

Answer 35

Explanation of Solution

As $n$ increases $Rmin$ decreases from the result of part (a),

$R≥ndn−1$ .

Here,

$R$ is the outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

The above expression can be written as,

$Rmin=ndn−1$

It is clear that as $n$ approaches to zero, $Rmin$ will tend to infinity.

Yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Conclusion:

Therefore, the value $Rmin$ tends to infinity and yes, as $n$ approaches to $1$ , the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Answer 36

(e)

Expert Solution

To determine

The value of

$Rmin$ .

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The value of

$Rmin$ .

Answer 41

Answer to Problem 35.49P

The value of

$Rmin$ is

$350 μm$ .

Answer 42

Explanation of Solution

Given info: The diameter of the fiber is $100 μm$ , and the index of refraction is $1.40$ .

Explanation:

The formula to calculate the $Rmin$ derived from part (a) is,

$Rmin=ndn−1$

Here,

$Rmin$ is minimum outside radius.

$n$ is the index of refraction.

$d$ is the diameter of the optical fiber.

Substitute $100 μm$ for $d$ and $1.40$ for $n$ in the above formula.

$Rmin=ndn−1=1.40×100 μm1.40−1=1.40×100 μm0.40=350 μm$

Conclusion:

Therefore, the value of $Rmin$ is $350 μm$ .

Answer 43

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Solution Summary: The author explains that the smallest outside radius is R_mathrmmin, which is allowed for a bend in the fiber if no light is to escape

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