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Chapter 33: Problem 24

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 between violet and red light can be calculated by performing the mentioned steps and simplifying the mathematical calculations.

Step by step solution

01

Identify Knowns

The given s are: Index of refraction \(n_v\) for violet light = 2.46, Index of refraction \(n_r\) for red light = 2.41, Angle of incidence \(i\) = 53.5掳, and Index of refraction of air \(n_{air}\) = 1.0 (it鈥檚 a standard value for air).
02

Convert Degrees to Radians

The angle of incidence should be converted from degrees to radians because trigonometric operations in the next steps require radian measures. To do so, use the conversion factor: \(1 radian = \frac{\pi}{180} degrees\). So, \(i\) in radians is \(53.5 \times \frac{\pi}{180}\).
03

Apply Snell's Law for Violet Light

We know that Snell's Law states \(n_{inc} \cdot \sin(i) = n_{ref} \cdot \sin(r)\). So, for the refraction angle of violet light \(r_v\), we can reformulate this equation to \(r_v = \sin ^{-1}(\frac {n_{inc}\sin(i)} {n_v})\). Since light is coming from air, the index of refraction for the incoming media \(n_{inc}\) is 1. After substituting \(i\) and \(n_v\) into the formula, we can calculate \(r_v\).
04

Apply Snell's Law for Red Light

Similarly, for the red light refraction angle \(r_r\), we get \(r_r = \sin ^{-1}(\frac {n_{inc}\sin(i)} {n_r})\). Substituting the given values gives the refraction angle for the red light.
05

Find the Angular Separation

The Angular separation \(螖r\) between the violet and red light can be calculated by finding the absolute difference of the refractive angles. So, \(螖r = |r_v - r_r|\). Calculate this difference to get the angular separation.

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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 bends when it passes from one medium to another. It is represented by the symbol \( n \). This value is a measure of how much the speed of light decreases in a material compared to its speed in a vacuum. The higher the index, the more the light slows down and consequently bends more. This bending happens due to changes in light speed when it travels from one medium into another.

For example, violet and red light have different indices of refraction in diamond, as given in the exercise: 2.46 for violet and 2.41 for red. These values indicate that violet light bends more than red light upon entering the diamond from the air, which has a default index of refraction of 1.0. This difference is because violet light has a shorter wavelength, causing it to interact more strongly with the atoms in the material. Understanding the index of refraction helps us predict how light will behave in different environments, and can be calculated using Snell's Law for determining the amount of light bending.
Angular Separation
Angular separation is the difference in the bending of light beams of different colors when they pass through a material. This is an important concept because it helps explain phenomena like rainbows and the dispersion of light through prisms. It is especially noticeable when light of different wavelengths refracts, since each color bends by a different amount.

In the exercise, we calculate the angular separation between violet and red light as they pass through diamond. This involves using Snell's Law to find the refraction angles for both colors of light. The angular separation \( \Delta r \) is then found by taking the absolute difference between these refraction angles, \( |r_v - r_r| \). By doing this, we can calculate how much violet and red light spread apart as they enter the diamond from the air, which leads to the separation of colors.
Wavelength-Dependent Refraction
Wavelength-dependent refraction is the reason why different colors of light refract or bend by different amounts when they pass through a medium. This occurs because each color corresponds to a different wavelength, and the amount a wave bends depends on its wavelength.

In materials like diamond, shorter wavelengths (such as violet light) are refracted more than longer wavelengths (such as red light). This is due to the fact that shorter wavelengths are more affected by the atomic structure of the material, leading to greater changes in direction. This behavior is scientifically known as dispersion.

This wavelength-dependent bending is what allows prisms to separate white light into its constituent colors, producing a spectrum. It is also why the index of refraction is used in calculations to determine how light of different wavelengths is bent differently. Thus, understanding wavelength-dependent refraction is essential for explaining natural color phenomena as well as for designing optical devices.

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Most popular questions from this chapter

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at \(23.0^{\circ}\) and \(62.0^{\circ},\) respectively, clockwise to that of the first. If unpolarized light is incident on the stack, the light has intensity \(55.0 \mathrm{~W} / \mathrm{cm}^{2}\) after it passes through the stack. If the incident intensity is kept constant but the second polarizer is removed, what is the intensity of the light after it has passed through the stack?

Birefringence is discussed in Section 33.5 and the refractive indexes for the two perpendicular polarization directions in calcite are given. A crystal of calcite serves as a quarter-wave plate; it converts linearly polarized light to circularly polarized light if the numbers of wavelengths within the crystal differ by one-fourth for the two polarization components. For light with wavelength \(589 \mathrm{nm}\) in air, what is the minimum thickness of a quarter-wave plate made of calcite?

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(\left(n_{2}=1.450\right)\) for the cladding and silica doped with germanium \(\left(n_{1}=1.465\right)\) for the core. (a) What is the critical angle \(\theta_{\text {crit }}\) 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 \(\theta_{\text {crit }}\) (Fig. \(\mathbf{P 3 3 . 4 6}\) ). Show that \(\sin \theta_{\mathrm{i}}=\sqrt{n_{1}^{2}-n_{2}^{2}}\). (c) What is the value of \(\theta_{\mathrm{i}}\) for \(n_{1}=1.465\) and \(n_{2}=1.450 ?\)

A beam of light strikes a sheet of glass at an angle of \(57.0^{\circ}\) with the normal in air. You observe that red light makes an angle of \(38.1^{\circ}\) with the normal in the glass, while violet light makes a \(36.7^{\circ}\) angle. (a) What are the indexes of refraction of this glass for these colors of light? (b) What are the speeds of red and violet light in the glass?

A layer of liquid sits on top of the horizontal surface of a transparent solid. For a ray traveling in the solid and incident on the interface of the two materials, the critical angle is \(38.7^{\circ}\). (a) For a ray traveling in the solid and reflecting at the interface with the liquid, for what incident angle with respect to the normal is the reflected ray \(100 \%\) polarized? (b) What is the polarizing angle if the ray is traveling in the liquid?
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