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Chapter 22: Problem 36

A beam of light is incident from air on the surface of a liquid. If the angle of incidence is \(30.0^{\circ}\) and the angle of refraction is \(22.0^{\circ}\), find the critical angle for the liquid when surrounded by air.

Short Answer

Expert verified
After calculating the refractive index of the liquid with air, the application of Snell's law gives the critical angle.

Step by step solution

01

Calculation of Refractive Index

Using Snell's law \( n1*sin(\Theta1) = n2*sin(\Theta2) \), where \( n1 \) and \( \Theta1 \) are the refractive index and angle of incidence for the first medium (air), \( n2 \) and \( \Theta2 \) are the refractive index and angle of refraction for the second medium (liquid). Since the refractive index of air is approximately 1, the formula becomes \( sin(\Theta1) = n2*sin(\Theta2) \), thus \( n2 = sin(\Theta1)/sin(\Theta2) \). Substituting the given angles, we get \( n2 = sin(30.0^{\circ})/sin(22.0^{\circ}) \)
02

Calculation of Critical Angle

The critical angle (\(\Theta_{c}\)) occurs when the refracted angle is at 90 degrees. Therefore, it can be calculated using the formula \( n2 = 1/sin(\Theta_{c}) \). Rearranging the equation, we get \( \Theta_{c} = arcsin(1/n2) \). Substituting the value of \( n2 \) obtained in Step 1, we get the critical angle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index is a measure of how much a ray of light bends, or refracts, when it enters a different medium. This property is crucial in understanding how light behaves in various substances. Each medium has a specific refractive index, which helps determine how light travels through it.
- In this context, the refractive index of air is approximately 1, indicating that light travels through air with minimal bending.
- By contrast, when light enters a liquid, its speed changes, and it refracts, or bends, depending on the refractive index of that liquid.
To find the refractive index of a liquid, Snell's Law is typically used. This involves applying the formula: \[ n_1 \sin(\Theta_1) = n_2 \sin(\Theta_2) \] where \( n_1 \) is for air and \( n_2 \) is for the liquid.
Angle of Incidence
The angle of incidence is the angle between the incoming light ray and the perpendicular, or normal, to the surface at the point of incidence. It is measured from the normal and is a key factor in determining how light will behave when encountering a boundary between two media.
- For this problem, the angle of incidence is given as \(30.0^{\circ}\). This is the angle at which the light initially strikes the liquid surface.
Understanding the angle of incidence helps explain why light changes direction when moving between different media, which is crucial for applying Snell's Law effectively.
Critical Angle
The critical angle is the angle of incidence that provides an angle of refraction of \(90^{\circ}\). Beyond this angle, all light is reflected back into the original medium, a phenomenon known as total internal reflection.
- To calculate the critical angle \( \Theta_c \), the refractive index of the liquid must be known.
Using the relationship: \[ \Theta_{c} = \arcsin\left( \frac{1}{n_2} \right) \] we substitute the refractive index calculated earlier to find the critical angle.
It's essential for understanding light behavior, especially in technologies relying on total internal reflection like fiber optics.
Angle of Refraction
The angle of refraction is the angle between the refracted ray and the normal. When light passes from one medium into another, such as from air to a liquid, this angle defines how much the path of the light is altered.
- In the problem, the angle of refraction is \(22.0^{\circ}\). This is how much the light bends once it enters the liquid.
The angle of refraction is calculated using Snell's Law, which helps predict how light will travel through different substances based on their refractive indices.

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Most popular questions from this chapter

Students allow a narrow beam of laser light to strike a water surface. They arrange to measure the angle of refraction for selected angles of incidence and record the data shown in the following table: $$ \begin{array}{cc} \hline \begin{array}{c} \text { Angle of Incidence } \\ \text { (degrees) } \end{array} & \begin{array}{c} \text { Angle of Refraction } \\ \text { (degrees) } \end{array} \\ \hline 10.0 & 7.5 \\ 20.0 & 15.1 \\ 30.0 & 22.3 \\ 40.0 & 28.7 \\ 50.0 & 35.2 \\ 60.0 & 40.3 \\ 70.0 & 45.3 \\ 80.0 & 47.7 \\ \hline \end{array} $$ Use the data to verify Snell's law of refraction by plotting the sine of the angle of incidence versus the sine of the angle of refraction. From the resulting plot, deduce the index of refraction of water.

The index of refraction for red light in water is \(1.331\) and that for blue light is \(1.340 .\) If a ray of white light enters the water at an angle of incidence of \(83.00^{\circ}\), what are the underwater angles of refraction for the blue and red components of the light?

A narrow beam of light is incident from air onto a glass surface with index of refraction \(1.56 .\) Find the angle ol incidence for which the corresponding angle of refraction is one-half the angle of incidence. Hint: You might want to use the trigonometric identity \(\sin 2 \theta=2 \sin \theta \cos \theta\).

(a) What is the energy in joules of an x-ray photon with wavelength \(1.00 \times 10^{-10} \mathrm{~m}\) ? (b) Convert the energy to electron volts. (c) If more penetrating \(x\) -rays are desired, should the wavelength be increased or decreased? (d) Should the frequency be increased or decreased?

A laser beam is incident at an angle of \(30.0^{\circ}\) to the vertical onto a solution of corn syrup in water. If the beam is refracted to \(19.24^{\circ}\) to the vertical, (a) what is the index of refraction of the syrup solution? Suppose the light is red, with wavelength \(632.8 \mathrm{~nm}\) in a vacuum. Find its (b) wavelength, (c) frequency, and (d) speed in the solution.
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