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Geometrical and Physical Optics

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Geometrical and Physical Optics

Did you know that roughly 90 million years ago our early ancestors used to have ultraviolet vision? As time went on, our eyesight shifted into the current ability to see the visual spectrum, which is the primary light we focus on in optics. Considering light's dual nature of behaving like a wave or a particle depending on the circumstances, two different field of optics developed. Geometric optics is a simple model of light which treats light consisting of rays that travel in straight lines, assuming that light travels in rays and can be reflected and refracted the boundaries between optical media. Physical optics, on the other hand, deals with the wave properties and phenomena of light such as interference and diffraction. In this article, we will establish the key differences between the two branches of optical physics!

Geometric and Physical Optics Similarities

As the title implies, geometric and physical optics both stem from the same branch of science - optical physics, but they describe different properties of light, and so as you might expect, the situations in which they are applied differ depending on the aspect of the physical system under consideration. To begin with, let's define what exactly optics is before delving into the specifics of each field.

Optics is the study of the behavior and characteristics of light.

Understanding how light behaves in different media and developing various instruments, such as mirrors, lenses, and interferometers, to further apply this knowledge to our everyday lives is a critical aspect of this physical science.

There are a number of different properties of light, but usually they can be categorized into two main types: particle phenomena and wave phenomena. Let's recall a critical property that light possesses - wave-particle duality.

Wave-particle duality states that the behavior of light can be described as either a particle or a wave.

Depending on the circumstances, sometimes it makes more sense to describe light as electromagnetic radiation that propagates through a medium in the form of a wave. Meanwhile, other phenomena, such as the photoelectric effect, can only be explained by considering light as a particle which we call a photon. So it makes sense to separate optics into two separate branches, focusing on different properties of light and with different applications.

Geometric and Physical Optics Differences

We have already established that besides the fact that geometric and physical optics deal with light, they will have varying applications. Let's look at each field separately so we can illuminate the distinction between the two closely related models of light phenomena.

Geometrical Optics

Geometric optics, also known as ray optics, is used when light waves interact with objects much greater than the wavelength of the visible light (\(400-700 \, \mathrm{nm}\)), therefore, utilizing the particle nature of light.

Geometrical optics are used when dealing with the transmission of light in rays.

An important aspect of geometrical optics is the image formation using light rays and devices such as lenses, mirrors, and prisms.

A ray of light is a hypothetical line representing a path along which light energy is transferred.

Rays realistically represent the observations made by scientists, but considering it is an abstraction, certain assumptions have to be made:

  1. In a consistent environment, the light rays travel in a straight line.

  2. If the light rays approach some optical aperture, they are stopped.

  3. Light rays can be absorbed, reflected or refracted.

  4. If light rays cross one another, no changes occur.

The main properties to focus on are reflection and refraction, whilst interference and diffraction are disregarded as we ignore the wave properties of light and instead are described using physical optics.

Physical Optics

In contrast to geometric optics, physical optics focuses on the wave nature of light.

Physical optics is used when dealing with the inherent nature and properties of light.

As a result, another term used to describe physical optics is wave optics.

A wave is a disturbance that moves through a medium or a vacuum and transmits energy.

In this context, light is a transverse electromagnetic (EM) wave in which the oscillations of the electric and magnetic field are perpendicular to the direction of travel of the wave. While mechanical waves such as sound require a medium to travel through, light is an EM wave; therefore it can travel through a vacuum as well, with a maximum speed of \(c=3\times10^8\, \frac{\mathrm{m}}{\mathrm{s}}\). The speed can be determined in other mediums, all we need to know is the frequency \(f\) and the wavelength \(\lambda\) of the wave.

The main properties studied by physical optics are interference, diffraction, and polarization, which we'll discuss in greater detail in the next section.

Geometrical and Physical Optics Applications

We already touched upon certain aspects each branch of optics focuses on, let's look at each of these concepts in more depth.

Geometrical Optics Applications

Imagine turning on a torch and pointing it at a flat mirror perpendicular to its surface. In this case, we'll be able to trace the overall path of the light as a straight line approaching the surface of the mirror and then reflecting off of it.

Reflection is the change in direction of a light ray, as it strikes a surface made up of a different medium to the medium it is initially travelling in.

So the light was initially traveling through air, and then bounced off of the mirror. Based on the law of reflection, the angle of reflection is always the same as the angle of incidence with respect to the normal, which is the line perpendicular to the surface. All of this occurs in the same plane.

Depending on the curvature of the mirror (i.e. flat, concave, or convex), the path of the rays will differ!

Now, let's say we direct the ray of light toward a still body of water. Some of the light will reflect off the surface, following the law of reflection stated previously. However, some of the light will transmit through the water. Water is much denser than air, so the path of the light will become redirected or refracted.

Refraction is the bending of light as it passes from one transparent medium into another.

The ability of certain materials to bend light is described using Snell's law and the refractive index \(n\), with the mathematical expression for both defined in the formula table later in the article.

The refractive index is a quantity associated with a specific optical medium and measures the amount of bending of light that occurs as the rays pass from one medium into another.

It's a dimensionless quantity, which depends on the velocity of light \(v\) in the specific medium. The more 'optically dense' the medium is, the more particles that get in the way of the light ray, resulting in a larger refractive index.

The refractive index for vacuum is \(n_\text{vacuum}=1\), while that of water is \(n_\text{water}=1.33\).

Both of these properties are visualized in Figure \(1\) below.

Geometrical and Physical Optics The reflection and refraction of a ray of light in two different mediums StudySmarterFig. 1 - The reflection and refraction of a ray of light as it coincides with the boundary between a less dense and more dense region, as described by Snell's law.

In this diagram, the incident ray of light is passing from a less dense medium \(n_1\) (e.g. air) into a more dense medium \(n_2\) (e.g. water).

Similarly, geometrical optics is used to analyze the rays of light interacting with various lenses, mirrors and prisms. There are two types of lenses and mirrors - concave and convex, with different properties and applications. Mathematically, they can be analyzed using the thin lens equation and the magnification equation, while graphically, we can determine the types of images formed in each case by drawing a ray diagram. More in-depth information on the various properties of different lenses is available in the thin lenses and the reflection in spherical surfaces explanations here on StudySmarter!

Physical Optics Applications

When light waves hit a bubble, we sometimes observe colorful patterns on it's surface. This is a result of interference, one of the properties studied in physical optics.

Interference occurs when two or more waves superimpose to form a new wave.

The resultant wave and its amplitude will depend on the relative phase of the waves. Constructive interference will occur if the waves meet in phase, and destructive interference if they meet out of phase. Mathematically, these can be expressed using a path difference equation, shown in the equation table below. Both of these occurrences are visualized in Figure \(2\) below. In these examples, the interfering waves have the same frequency so that the effects of constructive and destructive interference can be visualized more easily, but interference can occur between any two waves of different frequencies, amplitudes and relative phases.

Geometrical and Physical Optics Two waves in phase will superimpose into a wave with a larger amplitude in constructive interference, two waves out of phase will reduce the amplitude in destructive interference StudySmarterFig. 2 - Constructive interference occurs when waves are in phase with one another, whereas destructive interference results from waves out of phase.

Interference is demonstrated in Young's double slit experiment and is also demonstrated by diffraction gratings. As the light passed through two narrow slits, instead of seeing two bright spots corresponding to the slits, an interference pattern of waves overlapping and canceling one another is observed. The process of light encountering the slits and spreading out is known as diffraction.

Diffraction is the bending of light as it reaches an obstacle.

Diffraction will only occur if the slit width is similar in size to the wavelength of the light source. The equation used to mathematically describe a diffraction grating can be found in the formula table below.

Geometrical and Physical Optics Formula

All the most commonly used formulas in both, geometrical and physical optics, are compiled in the table below. For the purposes of this article, we will not go into too much detail about each equation. They will be discussed in more detail in other explanations here on StudySmarter.

NameEquationVariables
Snell's law$$n_1\sin\theta_1=n_2\sin\theta_2$$\(n_1\) - incident index\(\theta_1\) - incident angle\(n_2\) - refracted index\(\theta_2\) - refracted angle
Refractive index$$n=\frac{c}{v}$$\(n\) - index of refraction\(c\) - speed of light \(v\) - velocity in a substance
Thin lens equation $$\frac{1}{s_\mathrm{i}}+\frac{1}{s_\mathrm{o}}=\frac{1}{f}$$\(s_\mathrm{i}\) - image distance \(s_\mathrm{o}\) - object distance \(f\) - focal length
Magnification of a thin lens$$M=\frac{h_\mathrm{i}}{h_\mathrm{o}}=\frac{-s_\mathrm{i}}{s_\mathrm{o}}$$\(M\) - magnification\(h_\mathrm{i}\) - image height\(h_\mathrm{o}\) - object height\(s_\mathrm{i}\) - image distance\(s_\mathrm{o}\) - object distance
Wave equation$$ v = f \lambda $$\(v\) - velocity \(f\) - frequency\(\lambda\) - wavelength
Interference (double slit, diffraction grating)$$d\sin\theta=m\lambda$$\(d\) - separation\(\theta\) - angle\(m\) - an integer\(\lambda\) - wavelength
Path difference equation - constructive interference$$\Delta L=m\lambda$$\(\Delta L\) - distance \(m\) - an integer\(\lambda\) - wavelength
Path difference equation - destructive interference$$\Delta L=\left (m + \frac{1}{2} \right )\lambda$$\(\Delta L\) - distance \(m\) - an integer\(\lambda\) - wavelength

Example problems applying some of these formulas can be found later in this explanation.

Geometrical and Physical Optics Examples

Now that we've established what the main differences between geometric and physical optics are, let's apply this knowledge to some example problems!

One of the applications of geometric optics is obtaining and analyzing the images formed by thin lenses.

Firstly, let's look at the converging lens visible in Figure \(3\) below, and find the image this lens would form!

Geometrical and Physical Optics Converging lens ray diagram example setup StudySmarterFig. 3 - Geometrical optics applied to a converging lens.

Solution

To locate the image formed by this convex lens, we must draw a ray diagram. A minimum of two rays must be used to find their point of intersection, indicating the highest point of the image (the lowest point will be on the principal axis as that's where the object is rested). First of all we need to know some basic principles which we can follow which will tell us how to draw the path of light rays passing through the lens:

  1. Rays parallel to the principal axis will always pass through the lens in such a manner that they pass through the focus of the lens.
  2. Rays that are colinear with the principal axis will not be refracted by the lens as they are incident perpendicular to the surface of the lens. These rays pass in a smooth straight line, through the centre of the lens, intercepting the optical axis at a right-angle, before passing through the focus of the lens.

So, let's draw:

  1. Ray 1 parallel to the principal axis until it reaches the optical axis and then straight through the focus \(F\).
  2. Ray 2 passing straight through the center of the lens, where the optical and principal axis cross.

as visualized in Figure \(4\) below.

When a light ray passes through the center of a lens, it experiences no refraction; hence the ray travels in one smooth straight line. For the remaining rays, in this case, we chose to draw a parallel ray of light, which refracts at the optical axis and passes through the focus. However, we also could've drawn a ray going through the focus on the object side, where light would refract and continue parallel to the principal axis on the other side of the lens.

Geometrical and Physical Optics Converging lens ray diagram example with a ray going through the centre of the lens and a ray going through the opposite focus relative to the objectStudySmarterFig. 4 - The ray diagram of a converging lens.

This ray diagram provides us with a lot of information about the image obtained. For instance, we can see that it is flipped upside down or inverted, as well as appears smaller than the object itself, or in other words, it is diminished. Since the image is inverted, we can conclude that it is also real.

Real images can be projected onto a screen as they form on the opposite side of the object, while virtual images cannot as they are formed behind a lens.

All of these characteristics depend on the type of lens used and the location of the object regarding the focus of the lens. So, if this was a concave lens, the image would be virtual and upright.

Secondly, let's say that the object in Figure \(3\) is \(3\,\mathrm{cm}\) high and is located \(16.0 \, \mathrm{cm}\) away from the lens with a focal length of \(6.00 \, \mathrm{cm}\). Calculate the height of the image and how far away from the lens it forms!

Solution

To find the image distance \(s_\mathrm{o}\) we can use the following equation:

$$\frac{1}{s_\mathrm{i}}+\frac{1}{s_\mathrm{o}}=\frac{1}{f}.$$

By rearranging the thin lens equation and plugging in our values for object distance and focal length, we get

\begin{align} \frac{1}{s_\mathrm{i}}&=\frac{1}{f}-\frac{1}{s_\mathrm{o}} \\ \frac{1}{s_\mathrm{i}}&=\frac{1}{6.00 \, \mathrm{cm}}-\frac{1}{16.0 \, \mathrm{cm}} \\ \frac{1}{s_\mathrm{i}}&= 0.104\,\mathrm{cm} \\ {s_\mathrm{i}}&= 9.62 \,\mathrm{cm}. \end{align}

The positive value for image distance confirms our ray diagram results: the image is located on the opposite side of the lens and is inverted.

Now we can use the magnification equation

$$M=\frac{h_\mathrm{i}}{h_\mathrm{o}}=\frac{-s_\mathrm{i}}{s_\mathrm{o}}$$

to find the height of this image. We were given the height of the object and the object distance, and the image distance was calculated using the thin lens equation, so we get

\begin{align} h_\mathrm{i}&=\frac{-s_\mathrm{i} \, h_\mathrm{o}}{s_\mathrm{o}} \\ h_\mathrm{i}&= -\frac{(9.62 \, \mathrm{cm})(3 \, \mathrm{cm})}{(16.0 \, \mathrm{cm})} \\ h_\mathrm{i}&=-1.80 \, \mathrm{cm},\end{align}

where the negative sign implies that the image is inverted (as we established earlier using the ray diagram).

Now, let's look at an example problem applying the concepts of physical optics.

Monochromatic light with a wavelength of \(610 \, \mathrm{nm}\) falls normally on a diffraction grating with \(100\) lines per \(\mathrm{mm}\). Find the angle of the third order maximum.

Solution

First, we must calculate the separation distance \(d\) between the slits. We are told that there are \(N=100\) lines in every millimeter which converted to SI units is equivalent to \(10^{5}\) lines in every meter, so we can use the following equation

\begin{align} d&=\frac{1}{N} \\ d&=\frac{1}{10^5 \, \frac{\text{lines}}{\mathrm{m}}}=10^{-5} \, \mathrm{m}. \end{align}

To find the angle \(\theta\) of the third order (\(m=3\)) we rearrange the interference equation

$$ d\sin\theta=m\lambda$$

and plug in our values:

\begin{align} \sin\theta_3 &= \frac{m\lambda}{d} \\ \sin\theta_3 &= \frac{(3) (610\times 10^{-9} \, \cancel{\mathrm{m}})}{\left (10^{-5} \, \cancel{\mathrm{m}} \right )} \\ \sin\theta_3 &=0.183 \\ \theta_3 &= \sin^{-1}(0.183)=10.5^{\circ}. \end{align}

So, the angle of the third order maximum is equal to \(10.5^{\circ}\).

Geometrical and Physical Optics - Key takeaways

  • Optics is the study of the behavior and characteristics of light.
  • Wave-particle duality states that the behavior of light can be described as either a particle or a wave.
  • Geometric optics treats light as consisting of abstract rays which are perpendicular to the wavefronts of a wave and travel in straight lines in homogenous media.
  • Physical optics are used when dealing with the inherent nature and properties of light.
  • Some geometrical optics applications include explaining reflection, refraction, and behavior of light in lenses and mirrors.
  • Reflection is the change in direction of a light ray, as it strikes a surface made up of a different medium.
  • Refraction is the bending of light as it goes from one transparent medium into another.
  • Some physical optics applications are interference and diffraction.
  • Interference occurs when two or more waves superimpose to form a new wave.
  • Diffraction is the bending of light as it reaches an obstacle.

References

  1. Fig. 1 - The reflection and refraction of light in different mediums, StudySmarter Originals.
  2. Fig. 2 - Constructive and destructive interference, StudySmarter Originals.
  3. Fig. 3 - Converging lens ray diagram setup, StudySmarter Originals.
  4. Fig. 4 - Converging lens ray diagram, StudySmarter Originals.

Frequently Asked Questions about Geometrical and Physical Optics

Geometrical and physical optics is the study of behavior and characteristics of light, where geometrical optics explain and use various devices to form images, and physical optics focus on the properties of light. 

The applications of geometrical and physical optics are explaining the properties and behavior of light such as reflection, refraction, interference, and diffraction. 

The formulae for calculating geometrical and physical optics include the Snell's law, the lens equation, magnification of thin lenses, wave equation, and interference equations. 

The difference between physical and geometrical optics is that geometric optics are used when dealing with the transmission of light in rays, and physical optics are used when dealing with the inherent nature and properties of light.

The similarities between geometrical and physical optics are that they are both branches of optics studying the behavior and characteristics of light. 

Final Geometrical and Physical Optics Quiz

Question

What is the splitting of white light into a band of colors known as?

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Answer

Dispersion.

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Question

Which of the following is not an example of dispersion? 

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Answer

A hologram. 

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Question

Which of the following colors of light bends the least?

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Answer

Red.

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Question

What happens when a ray of white light enters a glass prism?

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Answer

The light spreads out. 

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Question

What is the mathematical expression for finding the refractive index of a medium? Here, \(n\) is the index of refraction, \(c\) is the speed of light, and \(v\) is the velocity in a substance.

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Answer

\(n=\frac{c}{v}\).

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Question

prism is a transparent body with four triangular surfaces inclined at an angle and three lateral surfaces. 

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Answer

True.

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Question

Modal dispersion causes white light to split into different colors, depending on the material's refractive index.

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Answer

False.

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Question

Which of the following is not a type of dispersion of light? 

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Answer

Optical dispersion.

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Question

To observe a rainbow, the Sun must be shining in front of the observer after the rain.

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Answer

False.

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Question

The bending of the light rays in the prism is determined by the law of refraction. What's the mathematical expression for this law? 

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Answer

\[n_1 \sin \theta_1 = n_2 \sin \theta_2,\] where \(n\) is the index of refraction and \(\theta\) is the angle of incidence or refraction.


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Question

What happens to a rainbow if the size of the droplets decreases? 

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Answer

Fewer colors are visible.

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Question

In a glass prism, as the wavelength increases, the refraction coefficient increases.

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Answer

True.

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Question

A ray of light travels through a vacuum and enters a glass prism. What is the speed of light in the prism, if it has a refractive index of \(1.52\)?

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Answer

\(1.97\times10^8\,\frac{\mathrm{m}}{\mathrm{s}}\).

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Question

What is the refractive index in a vacuum?

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Answer

\(1\).

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Question

Who was the first to demonstrate the scattering of light through a prism?

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Answer

Issac Newton.

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Question

The refraction index of a material is the ____ between the speed of light in vacuum and the speed of light in the material.

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Answer

ratio.

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Question

At the critical angle, what can we say about the refraction angle?

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Answer

The refraction angle is \(90^\circ\).

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Question

Which of the following equations describes the law of refraction, or Snell's law?

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Answer

\(n_1\sin\theta_1=n_2\sin\theta_2\).

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Question

When light travels from a medium with a low refraction index to one with a high refraction index, the refracted beam bends ____ the normal.

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Answer

toward.

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Question

When light travels from a medium with a high refraction index to one with a low refraction index, the refracted beam bends ____ the normal.

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Answer

away from. 

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Question

What occurs at any angle greater than the critical angle?

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Answer

Total internal reflection.

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Question

The change in direction (or bending) of a light ray when traveling from one medium to another is called ____. 

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Answer

refraction.

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Question

The propagation speed of light is ____ in materials with a higher refraction index. 

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Answer

slower.

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Question

Which of the following is the correct equation for the critical angle?

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Answer

\(\sin\theta_\mathrm{crit}=\frac{n_2}{n_1}\).

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Question

What causes a straw in a cup to appear bent at the surface of the liquid?

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Answer

Refraction.

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Question

The ratio of the refraction indices of two different media is ____ to the ratio of the light's propagation speed in each one.

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Answer

inversely proportional.

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Question

The speed of light is the same when traveling through all materials. 

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Answer

False.

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Question

What is the refractive index in vacuum?

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Answer

1.000.

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Question

What is the refractive index?

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Answer

The refractive index of a material is the ratio between the speed of light in vacuum and the speed of light in the material.

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Question

Which of the following is the correct equation for the refractive index?

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Answer

\(n=\frac{c}{v}\).

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Question

What are the units of the refractive index?

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Answer

The refractive index is dimensionless. 

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Question

The speed of light in a material with a high refractive index is ____ in a material with a low refractive index. 

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Answer

slower than. 

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Question

How does the critical angle depend on the indices of refraction?

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Answer

The critical angle is proportional to the inverse sine of the ratio between the indices of refraction.

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Question

When does total internal reflection occur?

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Answer

When a light ray traveling from a material of a greater refractive index to a smaller refractive index hits the surface at an angle greater than the critical angle. 

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Question

Which direction will a light beam traveling from glass to air bend? Assume the glass-air interface is smooth and that the light beam hits it at an angle. 

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Answer

Away from the normal to the surface. 

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Question

In which medium is the speed of light the greatest?

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Answer

air. 

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Question

How does the ratio of the refractive indices for two materials relate to the ratio of the speed of light in the materials?

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Answer

They are inversely proportional. 

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Question

What is the refracted angle for a light beam traveling from a material with a higher index of refraction to one with a lower one when the beam hits the surface at an angle greater than the critical angle?

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Answer

\(90^\circ\) with respect to the surface normal. 

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Question

A hand-held refractometer uses ___ to calculate the concentration of a liquid.

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Answer

refraction.

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Question

Which of the following statements is correct?

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Answer

A concave lens will always produce a virtual and upright image. 

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Question

Which of the following statements is correct?

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Answer

As the distance of an object to a concave lens increases, the distance of the image to the lens decreases.

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Question

What is the minimum number of rays required to sufficiently locate the image of an object in front of a thin lens?

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Answer

\(2\) rays.

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Question

If an object is located at the focal point of a convex lens, what kind of image will be formed? 

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Answer

Real, inverted, diminished.

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Question

Which equation can be used to find the magnification of a thin lens? 

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Answer

\(M = \frac{h_\mathrm{i}}{h_\mathrm{o}}\)

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Question

Converging lenses always have a negative focal length, meanwhile diverging lenses always have a positive focal length.

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Answer

False.

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Question

A candle of height \(6.50 \, \mathrm{cm}\) is located \(28.4 \, \mathrm{cm}\) away from a converging lens. If the focal length of this lens is \(15.2 \, \mathrm{cm}\), where is the image located?

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Answer

\(32.3 \, \mathrm{cm}\) on the opposite side of the lens.

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Question

A candle of height \(6.50 \, \mathrm{cm}\) is located \(28.4 \, \mathrm{cm}\) away from a diverging lens. If the focal length of this lens is \(15.2 \, \mathrm{cm}\), where is the image located?

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Answer

\(32.3 \, \mathrm{cm}\) on the opposite side of the lens.

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Question

What kind of image does a human eye produce?

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Answer

Real and inverted.

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Question

Which of the following statements describes the object distance \(s_\mathrm{o}\) most accurately?

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Answer

\(s_\mathrm{o}\) is always positive. 

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Question

Which conditions must be met for an image to be inverted?

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Answer

Negative magnification and image height.

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