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Chapter 33: Problem 22

The indexes of refraction for violet light \((\lambda = 400 \, \mathrm{nm})\) and red light \((\lambda = 700 \, \mathrm{nm})\) in diamond are 2.46 and 2.41, respectively. A ray of light traveling through air strikes the diamond surface at an angle of 53.5\(^\circ\) to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

Short Answer

Expert verified
The angular separation is approximately 0.69掳.

Step by step solution

01

Understand Snell's Law

Snell's Law helps us determine how light refracts when transitioning between two different media. The law is expressed as \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) and \( n_2 \) are the indices of refraction for the first and second medium respectively, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction relative to the normal.
02

Identify Known Values

We are given that \( n_{air} = 1.00 \), \( \theta_1 = 53.5^\circ \), \( n_{violet} = 2.46 \), and \( n_{red} = 2.41 \). These will be used in Snell's Law to calculate the angles of refraction for violet and red light.
03

Calculate Refraction Angle for Violet Light

Use Snell's Law for violet light: \( n_{air} \sin \theta_1 = n_{violet} \sin \theta_{violet} \). Substitute the known values: \( 1.00 \sin(53.5^\circ) = 2.46 \sin \theta_{violet} \). Solving for \( \theta_{violet} \), we find that \( \sin \theta_{violet} = \frac{\sin(53.5^\circ)}{2.46} \). Using a calculator, \( \theta_{violet} \approx 20.19^\circ \).
04

Calculate Refraction Angle for Red Light

Apply Snell's Law for red light: \( n_{air} \sin \theta_1 = n_{red} \sin \theta_{red} \). Substitute: \( 1.00 \sin(53.5^\circ) = 2.41 \sin \theta_{red} \). Solving for \( \theta_{red} \), we find \( \sin \theta_{red} = \frac{\sin(53.5^\circ)}{2.41} \). Thus, \( \theta_{red} \approx 20.88^\circ \).
05

Calculate Angular Separation

The angular separation \( \Delta \theta \) between the refracted violet and red light rays is given by the difference \( \Delta \theta = \theta_{red} - \theta_{violet} \). With \( \theta_{red} \approx 20.88^\circ \) and \( \theta_{violet} \approx 20.19^\circ \), we find \( \Delta \theta \approx 20.88^\circ - 20.19^\circ = 0.69^\circ \).

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Key Concepts

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

Index of Refraction
The index of refraction is a crucial concept in understanding how light behaves when it passes through different materials. It is represented by the symbol \( n \) and is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. Here's what you need to know:

  • The index of refraction for a medium is calculated as \( n = \frac{c}{v} \) where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.
  • A higher index of refraction indicates that light travels more slowly in that medium.
  • Different materials have different indices of refraction, which affect how light bends as it enters or exits the medium.

In the exercise, diamond has different indices of refraction for violet (2.46) and red (2.41) light, signifying that the speed of light in diamond is slower for violet light than for red light. Understanding these values helps us calculate the bending of light using Snell's Law.
Angular Separation
Angular separation refers to the difference in angle between two rays of light after they have passed through a medium and bent to varying extents. This concept is vital because it helps us understand dispersion, which is the process of separating light into its component colors.

Let's break down what causes angular separation and its significance:
  • When light enters a different medium, different wavelengths (colors) bend by different amounts. This is due to each having distinct indices of refraction.
  • The varying degrees of bending result in an angular spread that we observe as a spectrum.
  • In the exercise, we see that the violet and red light rays separate with an angular difference of \( 0.69^\circ \). This separation occurs because violet light refracts more due to its higher index of refraction compared to red light.

The angular separation is why we observe rainbows; each color bends differently as light passes through water droplets, creating a spectrum of colors.
Refraction Angle
The refraction angle is the angle formed between the refracted ray and the normal when light passes into a new medium. This angle tells us how much the light has bent and is calculated using Snell's Law.

The exercise illustrates this with the calculation of refraction angles for violet and red light:
  • For violet light, knowing the index of refraction is 2.46, we use Snell's Law to find the angle of refraction \( \theta_{violet} \approx 20.19^\circ \).
  • For red light, given its index of refraction is 2.41, we find the angle of refraction \( \theta_{red} \approx 20.88^\circ \).
  • By comparing these angles, students learn how different colors are refracted to different extents depending on their indices of refraction.

Understanding refraction angles allows us to predict and explain phenomena like the bending of light at glass surfaces, or in this case, the separation of colors in a diamond.

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Most popular questions from this chapter

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(n_2 = 1.4502\) for the cladding and silica doped with germanium \(n_1 = 1.4652\) for the core. (a) What is the critical angle \(\theta_{crit}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(theta_i\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{crit}\) (\(\textbf{Fig. P33.46}\)). Show that sin \(\theta_i\) =\( \sqrt {n^2_1 - n^2_2}\) . (c) What is the value of \(\theta_i\) for \(n_1\) = 1.465 and \(n_2\) = 1.450?

A beam of light is traveling inside a solid glass cube that has index of refraction 1.62. It strikes the surface of the cube from the inside. (a) If the cube is in air, at what minimum angle with the normal inside the glass will this light \(not\) enter the air at this surface? (b) What would be the minimum angle in part (a) if the cube were immersed in water?

A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction at this wavelength is 1.47? (b) What is the wavelength of these waves in the liquid?

A thin layer of ice (\(n\) = 1.309) floats on the surface of water (\(n\) = 1.333) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

A horizontal, parallelsided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of 35.0\(^\circ\) with the normal to the top surface of the glass. (a) What angle does the ray refracted into the water make with the normal to the surface? (b) What is the dependence of this angle on the refractive index of the glass?
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