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A narrow beam of light containing red (660 nm) and blue \((470 \mathrm{nm})\) wavelengths travels from air through a \(1.00-\mathrm{cm}\) -thick flat piece of crown glass and back to air again. The beam strikes at a \(30.0^{\circ}\) incident angle. (a) \(\mathrm{At}\) what angles do the two colors emerge? (b) By what distance are the red and blue separated when they emerge?

Short Answer

Expert verified
The red light emerges at an angle \(\theta_{r_{out}}\) and the blue light emerges at an angle \(\theta_{b_{out}}\), which can be found using Snell's Law in the steps described. The separation distance between the red and blue light upon emerging is \(\Delta d = d_r - d_b\), where \(d_r = 1.00 \thinspace cm \times \tan(\theta_{r_{out}})\) and \(d_b = 1.00 \thinspace cm \times \tan(\theta_{b_{out}})\).

Step by step solution

01

Compute the refraction angles using Snell's Law

First, we need to find the refractive indices of air, red light, and blue light in the crown glass. These values can be approximated as follows: - Refractive index of air: \(n_{air} \approx 1\) - Refractive index of crown glass for red light: \(n_r \approx 1.52\) (660 nm) - Refractive index of crown glass for blue light: \(n_b \approx 1.53\) (470 nm) Now, let's apply Snell's Law \(n_1 \times \sin (\theta_1) = n_2 \times \sin (\theta_2)\) to calculate the refraction angles \(\theta_r\) and \(\theta_b\) for red and blue light, respectively: For red light, \(1 \times \sin(30.0^\circ) = 1.52 \times \sin(\theta_r)\) For blue light, \(1 \times \sin(30.0^\circ) = 1.53 \times \sin(\theta_b)\) Now solve for \(\theta_r\) and \(\theta_b\).
02

Calculate the angles upon emerging from the glass

To calculate the emerging angles for red and blue light, apply Snell's Law again. This time, consider the light traveling from the glass to the air: For red light, \(1.52 \times \sin(\theta_r) = 1 \times \sin(\theta_{r_{out}})\) For blue light, \(1.53 \times \sin(\theta_b) = 1 \times \sin(\theta_{b_{out}})\) Solve for \(\theta_{r_{out}}\) and \(\theta_{b_{out}}\).
03

Calculate the separation between red and blue light upon emerging

To calculate the distance between the red and blue light beams upon emerging, first, find the difference in angles between the two beams: \(\Delta \theta = \theta_{r_{out}} - \theta_{b_{out}}\) Next, calculate the distance traveled by both beams along the horizontal axis. Since the thickness of the crown glass is 1.00 cm, we can express the distance as follows: \(d_r = 1.00 cm \times \tan(\theta_{r_{out}})\) \(d_b = 1.00 cm \times \tan(\theta_{b_{out}})\) Finally, find the difference in distance between the two beams: \(\Delta d = d_r - d_b\) The separation between the red and blue light beams upon emerging is \(\Delta d\).

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Key Concepts

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

Wavelength
Wavelength is a fundamental property of waves, including light waves. It is the distance between successive crests of a wave, typically measured in nanometers (nm) when dealing with light. In the electromagnetic spectrum, each color of light has a specific wavelength. For instance, red light has a longer wavelength, in this case, 660 nm, while blue light has a shorter wavelength, here 470 nm.

The wavelength of light determines its color and energy. Shorter wavelengths, like blue, have higher energy, whereas longer wavelengths, like red, have lower energy. In practical terms, when light travels through different mediums, such as air or glass, its speed changes due to its wavelength. This change in speed is partly responsible for phenomena like refraction, where light bends at the boundary between two different media.

Understanding wavelength is crucial when calculating how light behaves in different materials, particularly when applying principles like Snell's Law to determine angles of refraction.
Refractive Index
The refractive index is a measure of how much a substance can bend light. It is defined as the ratio of the speed of light in a vacuum to its speed in a given material. In mathematical terms, it is represented by the symbol \( n \). The refractive index indicates how much light is slowed down inside a material. For example:
  • Air has a refractive index of approximately 1, which means light travels almost at its maximum speed.
  • Red light in crown glass has a refractive index of about 1.52.
  • Blue light in crown glass has a refractive index of around 1.53.
When you compare these values, you can see that blue light is slowed a little more than red light in crown glass.

Refractive index is crucial for understanding refraction, where light bends when it enters a material with a different refractive index. This bending occurs due to the change in the speed of light. The larger the difference in refractive indices between two materials, the more the light will bend. This concept is central to Snell's Law, where you use these indices to calculate the angle of refraction when light passes from one medium to another.
Angle of Incidence
Angle of incidence is the angle formed between an incoming ray of light and the perpendicular (normal) to the surface at the point of incidence. It is a critical factor in determining how light refracts as it transitions between different media.

In our exercise, the angle of incidence is given as \(30.0^{\circ}\), which is the angle at which the beam of light hits the crown glass initially. The angle of incidence directly influences the angle of refraction based on Snell's Law, which states:
\[n_1 \times \sin(\theta_1) = n_2 \times \sin(\theta_2)\]
Here, \(\theta_1\) is the angle of incidence, and \(\theta_2\) is the angle of refraction.
  • A greater angle of incidence generally results in a larger angle of refraction.
  • If the angle of incidence is zero (light hits the surface head-on), there is no refraction.
Understanding the angle of incidence is crucial in predicting the behavior of light as it passes through different materials, aiding in calculations for optics applications.

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Most popular questions from this chapter

The most common type of mirage is an illusion that light from faraway objects is reflected by a pool of water that is not really there. Mirages are generally observed in deserts, when there is a hot layer of air near the ground. Given that the refractive index of air is lower for air at higher temperatures, explain how mirages can be formed.

(a) Light reflected at \(62.5^{\circ}\) from a gemstone in a ring is completely polarized. Can the gem be a diamond? (b) At what angle would the light be completely polarized if the gem was in water?

A light ray falls on the left face of a prism (see below) at the angle of incidence \(\theta\) for which the emerging beam has an angle of refraction \(\theta\) at the right face. Show that the index of refraction \(n\) of the glass prism is given by. $$n=\frac{\sin \frac{1}{2}(\alpha+\phi)}{\sin \frac{1}{2} \phi}$$ where \(\phi\) is the vertex angle of the prism and \(\alpha\) is the angle through which the beam has been deviated. If \(\alpha=37.0^{\circ}\) and the base angles of the prism are each \(50.0^{\circ},\) what is \(n ?\)

The light incident on polarizing sheet \(P_{1}\) is linearly polarized at an angle of \(30.0^{\circ}\) with respect to the transmission axis of \(\mathrm{P}_{1} .\) Sheet \(\mathrm{P}_{2}\) is placed so that its axis is parallel to the polarization axis of the incident light, that is, also at \(30.0^{\circ}\) with respect to \(\mathrm{P}_{1}\). (a) What fraction of the incident light passes through \(P_{1} ?\) (b) What fraction of the incident light is passed by the combination? (c) By rotating \(\mathrm{P}_{2},\) a maximum in transmitted intensity is obtained. What is the ratio of this maximum intensity to the intensity of transmitted light when \(P_{2}\) is at \(30.0^{\circ}\) with respect to \(\mathrm{P}_{1} ?\)

A light beam in air has an angle of incidence of \(35^{\circ}\) at the surface of a glass plate. What are the angles of reflection and refraction?

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