(II) For any type of wave that reaches a boundary beyond which its speed is increased, there is a maximum incident angle if there is to be a transmitted refracted wave. This maximum incident angle θiM corresponds to an angle of refraction equal to 90°. If θᵢ \u003e θiM, all the wave is reflected at the boundary and none is refracted, because this would correspond to sin θᵣ \u003e 1 (where is the angle θᵣ of refraction), which is impossible.(a) Find a formula for θiM using the law of refraction, Eq. 15–19.

Ch. 15 - Wave Motion

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo

Tutti i libri di testo

Giancoli Douglas 5th edition

Ch. 15 - Wave Motion

Problema 66a

Capitolo 15, Problema 66a

(II) For any type of wave that reaches a boundary beyond which its speed is increased, there is a maximum incident angle if there is to be a transmitted refracted wave. This maximum incident angle θ_iM corresponds to an angle of refraction equal to 90°. If θᵢ > θ_iM, all the wave is reflected at the boundary and none is refracted, because this would correspond to sin θᵣ > 1 (where is the angle θᵣ of refraction), which is impossible.
(a) Find a formula for θ_iM using the law of refraction, Eq. 15–19.

Guida verificata passo dopo passo

Step 1: Recall the law of refraction, also known as Snell's Law, which is expressed as:

n_{1} \sin θ i = n_{2} \sin θ r

. Here,

n_{1}

and

n_{2}

are the indices of refraction for the two media,

θ_{i}

is the angle of incidence, and

θ_{r}

is the angle of refraction.

Step 2: Recognize that the maximum incident angle

θ_{iM}

occurs when the angle of refraction

θ_{r}

is equal to 90°. Substitute

θ_{r}

= 90° into Snell's Law.

Step 3: Since

\sin (90)

is equal to 1, Snell's Law simplifies to:

n_{1} \sin θ_{iM} = n_{2}

. Rearrange this equation to solve for

θ_{iM}

Step 4: Divide both sides of the equation by

n_{1}

to isolate

\sin θ_{iM}

\sin θ_{iM} = \frac{n_{2}}{n_{1}}

Step 5: To find the maximum incident angle

θ_{iM}

, take the inverse sine (arcsin) of both sides:

θ_{iM} = \arcsin (\frac{n_{2}}{n_{1}})

. This is the formula for the maximum incident angle.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Law of Refraction (Snell's Law)

The Law of Refraction, also known as Snell's Law, describes how waves change direction when they pass from one medium to another. It is mathematically expressed as n₁ sin(θᵢ) = n₂ sin(θᵣ), where n₁ and n₂ are the refractive indices of the two media, and θᵢ and θᵣ are the angles of incidence and refraction, respectively. This law is fundamental in understanding how light and other waves behave at boundaries.

Video consigliato:

Percorso guidato

07:59

Snell's Law

Critical Angle

The critical angle is the angle of incidence above which total internal reflection occurs, meaning that all the incident wave is reflected back into the original medium. It is defined as θ_c = arcsin(n₂/n₁) when n₁ > n₂. When the angle of incidence exceeds this critical angle, the refracted wave cannot exist, leading to the phenomenon described in the question.

Video consigliato:

Percorso guidato

01:04

Critical Angle

Total Internal Reflection

Total internal reflection is a phenomenon that occurs when a wave traveling in a medium hits a boundary with a less dense medium at an angle greater than the critical angle. In this case, the wave is completely reflected back into the original medium, and no refraction occurs. This principle is crucial in applications such as fiber optics, where light is kept within the fiber by repeated total internal reflections.

Video consigliato:

Percorso guidato

05:29

Total Internal Reflection

Pratica correlata

Domanda del libro di testo

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 45-cm-wide tub is 0.85 Hz. What is the speed of the water wave?

1054

views

Domanda del libro di testo

A guitar string is supposed to vibrate at 247 Hz, but is measured to actually vibrate at 262 Hz. By what percentage should the tension in the string be changed to get the frequency to the correct value?

1209

views

Domanda del libro di testo

A bug on the surface of a pond is observed to move up and down a total vertical distance of 0.10 m, lowest to highest point, as a wave passes. If the amplitude increases to 0.15 m, by what factor does the bug’s maximum kinetic energy change?

945

views

Domanda del libro di testo

A particular violin string plays at a frequency of 294 Hz. If the tension is increased 22%, what will the new frequency be?

1141

views

Domanda del libro di testo

A longitudinal earthquake wave strikes a boundary between two types of rock at a 41° angle. As the wave crosses the boundary, the specific gravity of the rock changes from 3.6 to 2.8. Assuming that the elastic modulus (Section 15–2)is the same for both types of rock, determine the angle of refraction.

1263

views