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Chapter 9: Problem 15

Show that a plane wave normally incident on one face of a wedge-shaped prism of angle \(A\) is deviated by an amout \(\theta_{\text {dev }}\), where $$ n \sin A=\sin \left(A+\theta_{\text {dev }}\right) \text { . } $$

Short Answer

Expert verified
The problem relationship is verified using Snell's Law for normal incidence.

Step by step solution

01

Understand the Problem

We need to show that a plane wave, when incident normally on a wedge-shaped prism with an angle \(A\), is deviated by an angle \(\theta_{\text{dev}}\). The relationship given is that \(n \sin A = \sin(A + \theta_{\text{dev}})\), where \(n\) is the refractive index of the medium of the prism compared to the outside medium.
02

Snell's Law and Normal Incidence

At the first interface (incidence on the prism), since the wave is normally incident, the angle of incidence \(i = 0\) and there is no refraction involved. This means the light continues straight into the prism without bending. Snell's Law, which states \(n_1 \sin i = n_2 \sin r\), simplifies to \(0 = n_2 \sin r\), implying \(r = 0\).
03

Refraction at Second Interface

Upon reaching the second interface of the prism, the light refracts out into the air. Here the angle of incidence is \(A\) because of the wedge angle. By Snell's Law applied at this interface, we have \(n \sin A = \sin r'\), where \(r'\) is the angle of refraction. This tells us that \(r'\) is related to the deviation angle \(\theta_{\text{dev}}\).
04

Relate Angle of Deviation

The deviation angle \(\theta_{\text{dev}}\) is the difference between the final direction of the light and its initial path. Since \(r'\) is the angle into the air, it follows that \(\theta_{\text{dev}} = r' - A\). From the equation \(n \sin A = \sin r'\), substitute \(r' = A + \theta_{\text{dev}}\).
05

Verify the Given Relationship

Substitute \(r' = A + \theta_{\text{dev}}\) into the equation \(n \sin A = \sin r'\) to get \(n \sin A = \sin(A + \theta_{\text{dev}})\). This verifies the relationship provided in the problem.

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Key Concepts

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

Snell's Law
Snell's Law is a fundamental principle in optics that describes how light bends or refracts as it passes from one medium to another. When light encounters a boundary between two different optical mediums, its speed changes, causing it to change direction. The law is mathematically represented as: \[ n_1 \sin i = n_2 \sin r \] Here, \(n_1\) and \(n_2\) are the refractive indices of the initial and secondary medium, respectively, \(i\) is the angle of incidence, and \(r\) is the angle of refraction.
At the heart of Snell's Law is the conservation of energy and momentum in the plane of the interface.
If a light wave enters a medium at an angle (not perpendicular), it bends either towards or away from the normal, depending on whether the second medium is denser or less dense than the first. Understanding Snell's Law is crucial for predicting how waves behave at boundaries, especially in devices like prisms where light enters and exits at angles.
  • This principle is used in a variety of applications, from designing lenses and fiber optics to explaining natural phenomena like rainbows.
  • Snell's law is instrumental in this problem because it helps explain the angles of incidence and refraction as the wave travels through the wedge-shaped prism.
Refractive Index
The refractive index is a dimensionless number that indicates how much light or any other wave is bent, or refracted, when entering a material.
It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. This can be expressed 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.
The refractive index determines how much the path of light is bent, or refracted, when entering the material.
  • A higher refractive index indicates that the light will slow down more and bend at a greater angle.
  • In our problem, the refractive index \(n\) of the prism's medium is crucial in determining the deviation angle \(\theta_{\text{dev}}\).
Knowing the refractive index allows us to apply Snell's Law to calculate angles and understand how light will behave as it moves between two mediums.
It is essential in designing lenses and other optical devices, as it requires precise calculations of how light is focused and directed.
Wedge-shaped prism
A wedge-shaped prism is an optical device that is often used to disperse light or to change its direction.
It has at least one angle that significantly differs from 90 degrees, called the apex angle \(A\).
When a light wave hits the prism, it refracts at the two surfaces at different angles due to the non-parallel nature of the surfaces.In the exercise, we're dealing with a wave entering the prism normally (perpendicularly), so initially, there is no deviation.
As light travels inside the wedge-shaped prism and reaches the second surface, it is refracted again and deviates by an angle \(\theta_{\text{dev}}\).
  • The wedge shape is significant because the non-parallel sides cause light to bend, which is described by Snell's Law.
  • This bending leads to a deviation of the light's path as it exits, creating a measurable angle.
Understanding how wedge-shaped prisms work helps grasp how light dispersion and deviation occur, which are fundamental concepts in optics.
This knowledge is crucial for applications in fields like spectroscopy, where precise light deviation and separation into components are necessary.

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Most popular questions from this chapter

Use a pinhole in a piece of aluminum foil illuminated by a broad source to study the floaters. When you are looking at one, try to wipe it off by blinking. Can it be wiped? "Roll" your eyes once or twice and then watch the floaters swirl! Now try to find if they are nearer to the pupil or to the retina: vary the distance of the pinhole from your eye. The size of the circle of light changes. (Make a sketch to aid in explaining why.) Any object at the same location as the pupil would change apparent size by the same ratio as does the projection of the pupil. (Why?) Anything at the retina (or near it) would not change apparent size at all. (Why?) What do the floaters do? Are they closer to the retina or to the pupil? Now try to estimate their length and diameter. To estimate the diameter, compare them with a human hair (yours) held in front of your pupil between the pinhole and the pupil. For this purpose, you needa very small pinhole, smaller than you can easily make with a pin. Crinkle up a piece of aluminum foil, then smooth it. Look for an accidental small pinhole. (You can tell if it is small -less light intensity gets through.) Now look at a hair. You should see its shadow and see nice diffraction fringes at its edge. Compare its size with that of a floater. Are they finer than a hair? (Note: A human hair has a diameter about \(\frac{1}{20} \mathrm{~mm}\), i.e., \(50 \mu\) (micron). A typical red blood cell has a diameter of 5 or \(6 \mu\).)

A "corner reflector" consists of three plane mirrors joined so as to form an inside corner of a rectangular box. Show that a light beam that strikes a corner reflector is directed back at 180 deg to its original direction, independent of the angle of incidence, as long as it hits all three surfaces.

When you put your face under water and try looking without a face mask, everything looks blurred, because the change of index of refraction in going from water to eye is not very great. As a simplification, assume there is no change in index. Also assume your eye lens has very little effect, as if all the focusing were done at the first air-to-eye interface. (This is a crude approximation. Actually, you can see underwater to same extent.) Assume the focal length of that first surface is \(3 \mathrm{~cm}\), and that a parallel beam of light in air is brought to a focus at the retina. When you look underwater, you lose that focusing action. Design glasses that can be worn underwater so as to enable you to see clearly. Use glass with index of refraction 1.5. Show that if the focal length when used underwater is \(3 \mathrm{~cm}\), then the focal length when used in air is about \(1 \mathrm{~cm} .\) If one of these glass lenses is used as an ordinary magnifying glass what is its magnification? Suppose you use an ordinary glass marble for the lens. You want to form an image (of a parallel beam in water) \(3 \mathrm{~cm}\) behind the rear surface of the marble. What should be the diameter of the marble?

The world's cheapest broad, almost monochro- matic light source is obtained by burning a wad of toilet paper. You can use this source to see Fabry-Perot fringes. Burn the paper. (The room should be dark-perhaps also you should have some water handy!) Look through the flame at the image of the flame at near- normal incidence in a piece of glass-a microscope slide or picture- frame glass. You will see finger-print-like fringes. If the glass is optically flat, the fringes will be circles centered on your eyeballs; in any case, you can see them easily. If you have a gas stove or bunsen burner, you can get a brighter monochromatic sodium source by sprinkling salt on a wet knife and immersing it in the flame. Then you can see the Fabry-Perot fringes even in the daytime. For a nice, steady, broad monochromatic source with which to look at the fringes, use the G.E. neon bulb NE-34.

Suppose you cover one of two slits with a microscope slide and the other with nothing. If the slide has thickness \(1 \mathrm{~mm}\), show that monochromatic light of wavelength \(5000 \AA\) gets a retardation in one slit of about 1000 wavelengths relative to the other. If the double-slit pattern is not to wash out, the light must be fairly monochromatic. How narrow a band of wavelengths (in angstroms) is required so that the relative phase shift of the two slits varies by less than 180 deg from one edge of the wavelength band to the other? How could you use this fact to measure the bandwidth of a spectral line? (What would you measure and plot versus what, and how would you obtain the bandwidth from the plot?)
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