Problem 48
Question
The angle of incidence and angle of refraction along a particular interface between two media are \(30^{\circ}\) and \(40^{\circ}\), respectively. What is the critical angle for the same interface?
Step-by-Step Solution
Verified Answer
The critical angle is approximately \(53.13^{\circ}\).
1Step 1: Identify the Known Quantities
We are given the angle of incidence, \(\theta_i = 30^{\circ}\), and the angle of refraction, \(\theta_r = 40^{\circ}\), for an interface between two media. We aim to find the critical angle.
2Step 2: Apply Snell's Law
Snell's Law relates the angles of incidence and refraction to the refractive indices of the two media: \(n_1 \sin(\theta_i) = n_2 \sin(\theta_r)\).
3Step 3: Calculate Unknowns Using Snell's Law
Solving Snell's Law for one side, we get \(\frac{n_1}{n_2} = \frac{\sin(\theta_r)}{\sin(\theta_i)} = \frac{\sin(40^{\circ})}{\sin(30^{\circ})}\). Calculate this value to determine the ratio of refractive indices \(\frac{n_1}{n_2}\).
4Step 4: Determine the Critical Angle
The critical angle \(\theta_c\) occurs when the angle of refraction is \(90^{\circ}\). Thus, using Snell's Law for the critical angle: \(n_1 \sin(\theta_c) = n_2 \sin(90^{\circ})\). This simplifies to \(\sin(\theta_c) = \frac{n_2}{n_1}\). Calculating \(\sin(\theta_c)\) from \(\frac{n_1}{n_2}\) found earlier, we find the critical angle by taking \(\theta_c = \arcsin\left(\frac{n_2}{n_1}\right)\).
5Step 5: Calculate the Critical Angle
Using the earlier value, \(\frac{n_1}{n_2} = \frac{\sin(40^{\circ})}{\sin(30^{\circ})}\), calculate \(\sin(\theta_c) = \frac{n_2}{n_1} = \frac{1}{\frac{n_1}{n_2}}\). Then, \(\theta_c = \arcsin\left(\sin(\theta_c)\right)\) to determine the critical angle.
Key Concepts
Snell's LawAngle of IncidenceAngle of RefractionRefractive Indices
Snell's Law
Snell's Law is a fundamental principle that explains how light bends, or refracts, when it travels from one transparent medium into another. It describes the relationship between the angles of incidence (where the light enters) and refraction (where the light exits), as well as the refractive indices of the two materials. The formula for Snell's Law is given by:\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_r) \]Here:
- \( n_1 \) is the refractive index of the first medium.
- \( n_2 \) is the refractive index of the second medium.
- \( \theta_i \) is the angle of incidence.
- \( \theta_r \) is the angle of refraction.
Angle of Incidence
The angle of incidence is the angle formed by a ray or wave striking a surface, measured from a normal line that is perpendicular to that surface. In simpler terms, it is the angle at which the incoming light hits the boundary between two different media, like air and water.
Key points about the angle of incidence include:
Key points about the angle of incidence include:
- It is usually denoted as \( \theta_i \).
- In the given problem, the angle of incidence is \( 30^{\circ} \).
- This angle is crucial as it affects the behavior of the light at the boundary, dictating how much the light will bend or refract.
Angle of Refraction
The angle of refraction is the angle that a refracted wave or ray makes with the normal after passing through the interface between two media. Just like the angle of incidence is the angle at which light arrives, the angle of refraction is the angle at which light bends as it enters a second medium.
Important points regarding the angle of refraction:
Important points regarding the angle of refraction:
- It is denoted by \( \theta_r \).
- In the example, the light refracts at an angle of \( 40^{\circ} \).
- This angle changes depending on the refractive indices of the two materials.
Refractive Indices
Refractive indices are numerical values that describe how much light will bend when it enters a different medium. Each material has a characteristic refractive index, which indicates how much slower light travels through the medium in comparison to a vacuum.
Key details about refractive indices include:
Key details about refractive indices include:
- The refractive index is denoted by \( n \).
- It affects how light refracts when moving between two media.
- In the problem, the ratio of the refractive indices \( \frac{n_1}{n_2} \) was used to determine the bending of light.
Other exercises in this chapter
Problem 44
The index of refraction of crown glass is 1.515 for red light and 1.523 for blue light. (a) If light of both colors is incident on crown glass from air, the blu
View solution Problem 45
A beam of light with red and blue components of wavelengths \(670 \mathrm{nm}\) and \(425 \mathrm{nm}\), respectively, strikes a slab of fused quartz at an inci
View solution Problem 49
Light passes from medium A into medium \(B\) at an angle of incidence of \(30^{\circ} .\) The index of refraction of A is 1.5 times that of B. (a) What is the a
View solution Problem 53
An optic fiber of index of refraction of 1.6 is surrounded by a layer called the cladding layer. The index of refraction of the cladding is \(1.3 .\) If the ang
View solution