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Chapter 23: Problem 51

The indices 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 of red and violet light in diamond is calculated using Snell's Law for each color.

Step by step solution

01

Understanding Snell's Law

Snell's Law relates the angle of incidence to the angle of refraction for light passing between two media, and is given by the formula:\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]where \(n_1\) and \(n_2\) are the indices of refraction for air and diamond, respectively. \(\theta_1\) is the angle of incidence, and \(\theta_2\) is the angle of refraction.
02

Set Known Values

For this problem:- The index of refraction for air, \(n_1\) is approximately 1.- The angle of incidence, \(\theta_1\), is given as \(53.5^\circ\).- The index of refraction for violet light, \(n_2\), is 2.46.- The index of refraction for red light, \(n_2\), is 2.41.
03

Calculate the Angle of Refraction for Violet Light

Using Snell's Law for violet light:\[ 1 \cdot \sin(53.5^\circ) = 2.46 \cdot \sin(\theta_{v}) \]Solve for \(\theta_{v}\) (angle of refraction for violet light):\[ \sin(\theta_{v}) = \frac{\sin(53.5^\circ)}{2.46} \]\[ \theta_{v} = \arcsin\left(\frac{\sin(53.5^\circ)}{2.46}\right) \]Calculate \(\theta_{v}\) using a calculator.
04

Calculate the Angle of Refraction for Red Light

Similarly, using Snell's Law for red light:\[ 1 \cdot \sin(53.5^\circ) = 2.41 \cdot \sin(\theta_{r}) \]Solve for \(\theta_{r}\) (angle of refraction for red light):\[ \sin(\theta_{r}) = \frac{\sin(53.5^\circ)}{2.41} \]\[ \theta_{r} = \arcsin\left(\frac{\sin(53.5^\circ)}{2.41}\right) \]Calculate \(\theta_{r}\) using a calculator.
05

Calculate the Angular Separation Between the Two Colors

The angular separation \(\Delta \theta\) between the refracted violet and red light is given by:\[ \Delta \theta = \theta_{r} - \theta_{v} \]Subtract the calculated \(\theta_{v}\) from \(\theta_{r}\) to find \(\Delta \theta\).

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Key Concepts

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

Refraction
When light passes from one medium to another, like from air into diamond, the speed of light changes. This change in speed causes the light to bend, a process known as refraction. The bending occurs because different materials have distinct optical densities. The extent to which light bends when entering a new medium is determined by its index of refraction. Refraction is governed by Snell's Law, which relates the angles and indices of refraction of the two media involved. In the given exercise, we are observing how light bends when it enters diamond from the air. This bending effect can differ based on the light's wavelength, leading to the separation of colors.
Light Wavelength
Light is composed of waves, and each color of light has a characteristic wavelength. In our exercise, we are dealing with violet light, which has a wavelength of 400 nm, and red light, with a wavelength of 700 nm. The wavelength of light influences how much it refracts when passing through different media. Shorter wavelengths tend to bend more than longer ones, which is why violet light refracts more than red light in diamond. This difference in refraction based on wavelength leads to the phenomenon of dispersion, which is the basis for observing angular separation between different colors in the refracted light.
Angular Separation
Angular separation refers to the difference in direction between two rays of light after they have been refracted. In our example, it pertains to the separation between the paths of violet and red light as they emerge into the diamond. This occurs because each color light ray bends at a different angle due to its specific wavelength. Angular separation is calculated using the angles of refraction for each color, which are determined by Snell's Law calculations. The greater the difference in wavelengths, the more noticeable the angular separation. This is why prisms can split white light into its constituent rainbow colors.
Index of Refraction
The index of refraction, denoted as 'n', is a measure of how much a light wave slows down when entering a material compared to its speed in a vacuum. Each material has its own index, which affects how much light is refracted. In this task, diamond's indices of refraction for violet (2.46) and red light (2.41) tell us how much each color will bend. A higher index indicates that light travels slower in that medium, bending more. Thus, violet light, with its higher index, refracts more than red light, causing a noticeable angular separation when both colors pass through diamond.

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Most popular questions from this chapter

The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

Laser surgery. Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portion back into place. In one such procedure, a laser beam has a wavelength of \(810 \mathrm{nm}\) and delivers \(250 \mathrm{~mW}\) of power spread over a circular spot \(510 \mu \mathrm{m}\) in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of \(1.34 .\) (a) If the laser pulses are each \(1.50 \mathrm{~ms}\) long, how much energy is delivered to the retina with each pulse? (b) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is \(30^{\circ}\), the intensity of the emerging beam is \(I\). If you instead want the intensity to be \(I / 2,\) what should be the angle between the filter axis and the original direction of polarization of the light?

Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

A radio broadcasts at a wavelength of \(3.06 \mathrm{~m}\). Near the broadcast tower the electric-field amplitude is \(2700 \mathrm{~N} / \mathrm{C}\). Calculate (a) the frequency, (b) the wave number, and (c) the magnetic-field amplitude of the electromagnetic wave.
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