PHYS102 Study Guide

Unit 7: Optics

7a. Determine the size, location, and nature of images by using the mirror and lens equations

How do mirrors form images?
What is the primary difference between concave and convex mirrors?
What are the two possible types of images? How is the type of image formed by a mirror determined by its location relative to the mirror?
How are focal length and curvature of a mirror related?
How are the locations of an object and its image related by the mirror equation?
How do lenses form images?
What is the primary difference between converging and diverging lenses?
How is the type of an image formed by a lens determined by its location relative to the lens?
How are the locations of an object and its image related by the lens equation?
What is magnification, and what does its sign (positive/negative) indicate about the image?

Mirrors form images of objects by reflection. As the rays of light that an object generates bounce off the reflective surface of a mirror, they intersect and form an image. In specular reflection, the angle between the reflected ray and the normal of the surface is equal to the angle of incidence between the incident ray and the normal of the surface. When the surface is rough or dirty, light rays do not get reflected in this simple manner, and the surface appears dull because the rays do not reproduce an image. This is called diffuse reflection.

A concave parabolic surface (a concave mirror) has the property of reflecting all rays parallel to its axis of symmetry in such a way that they all intersect at one point: the focal point of the parabola.

Spherical surfaces serve as a good replacement for parabolic ones; as long as the incident rays fall near the axis of symmetry, their reflections also intersect at one focal point that is located on the axis of symmetry halfway between the center of curvature and the mirror. Thus, the focal length $f$ of the spherical mirror and its radius of curvature $R$ are related as $f=\frac{R}{2}$ .

Images formed by the intersection of rays reflected back toward the object are real images. If the reflected rays do not intersect, then their extensions intersect behind the mirror, forming a virtual image. A virtual image can be seen, but cannot be captured with a camera film or detector.

When describing the locations of objects and images formed by mirrors, the following conventions are used:

Positions in front of a mirror are considered positive. The distance between an object and the mirror is denoted by $d_{o}$ and is always positive, as the object is always located in front of the mirror. The location of an image is denoted by $d_{i}$ , and is positive when the image is in front of the mirror and negative when the image is behind the mirror. Thus, $d_{i}$ is positive for real images and negative for virtual images.

We define magnification as the ratio of the sizes of the image and the object: $M=\frac{h_{i}}{h_{o}}$ . Here, $h_{i}$ is the height of the image, and $h_{o}$ is the height of the object. If the image is upside down (or inverted) then $h_{i}$ is considered negative, which means $M$ is also negative. A virtual image is always upright; a real image is always inverted.

We can use three rays to draw the images formed by mirrors:

The reflection of the rays parallel to the axis of symmetry pass through the focal point
The reflection of the rays passing through the focal point are parallel to the axis of symmetry
The rays perpendicular to the surface (or passing through the center of curvature, for spherical surfaces) are reflected along the same line

Plane mirrors have an infinite radius of curvature, and thus no focal point. The image of an object located in front of a mirror is located at the same distance behind the mirror. This image is virtual, upright, and has the same size as the object.

Concave mirrors have a reflective surface on the inside of a spherical surface, so their focal point is in front of the mirror and on the same side as an object. Their focal length is positive: f > 0. Concave mirrors can form either real or virtual images, depending on the location of the object.

Convex mirrors have a reflective surface on the outside of a spherical surface, so their focal point is behind the mirror. Their focal length is negative: f < 0. Concave mirrors always form virtual images.

If we know the location $p$ and size $h_{o}$ of the object, we can use the mirror equation: $\frac{1}{d_{o}}+\frac{1}{d_{i}}=\frac{1}{f}$ to find the location of the image. We can use the magnification formula $M=-\frac{d_{i}}{d_{o}}=\frac{h_{i}}{h_{o}}$ to find the size of the image.

Lenses form images of objects by refraction. As rays of light change direction when crossing into or out of the material of the lens (typically glass or plastic), they intersect and form an image. The lens has to be very thin compared to the distance between the lens and object. Lenses are made in a variety of shapes, and they have focal points where the refracted rays converge if the incident rays are parallel to the lens' axis of symmetry. Lenses can be made to be either converging or diverging.

For lenses, the object and image locations are similar to that of mirrors. The distance between an object and the lens is denoted by $d_{o}$ , which is always positive. However, real images are located on the opposite side (behind the lens), so the distance between the image and the lens, $d_{i}$ , is positive is when the image is behind the lens (a real image) and negative when the image is in front of the lens (a virtual image)

For lenses, magnification is defined as with mirrors, as the ratio of the sizes of the image and the object: $M=\frac{h_{i}}{h_{0}}$ . Here, $h_{i}$ is the height of the image and $h_{0}$ is the height of the object.

If the image is upside down (or inverted) then $h_{i}$ is considered negative, which means $M$ is also negative. From geometric considerations, it can be shown that $M=-\frac{d_{i}}{d_{0}}$ . From here, it follows that $M$ is positive when $d_{i}$ is negative, so the virtual image is always upright, and that $M$ is negative when $d_{i}$ is positive, so the real image is always inverted.

Converging lenses refract incident rays so that they bend toward the axis of symmetry, while diverging lenses bend incident rays away from the axis of symmetry. There are three rays that can be used to draw the images formed by the lenses:

The incident rays parallel to the axis of symmetry refract so that the refracted rays pass through the focal point behind the converging lens, or so that their extensions pass through the focal point in front of the diverging lens
The incident rays passing through the focal point refract so that they are parallel to the axis of symmetry
The rays passing through the center of the lens are not refracted

Converging lenses have a positive focal length and can form real and virtual images, depending on the location of the object. Diverging lenses have a negative focal length and always form virtual images.

If the location $d_{0}$ and size $h_{0}$ of the object are known, the location of the image can be found from the lens equation: $\frac{1}{d_{0}}+\frac{1}{d_{i}}=\frac{1}{f}$ . Then, the size of the image can be found using the magnification formula: $M=-\frac{d_{i}}{d_{0}}=\frac{h_{i}}{h_{0}}$ .

Review The Ray Aspect of Light, Image Formation by Mirrors, and the corresponding rules for lenses in Image Formation by Lenses.

7b. Solve problems using the law of refraction

How are the angle of incidence and angle of refraction related?
When does total internal reflection occur? What is a critical angle?

Refraction is the change of direction, or bending, of a light ray when it crosses a boundary between two media. It occurs because the light propagates at different speeds in media with different optical properties, which are determined by the electric permittivity and the magnetic permeability of the media. The speed of light in vacuum is $c=3\times 10^{8}$ m/s. In other media, it is decreased by a factor of $n$ , the index of refraction of the medium: $v_{light}=\frac{c}{n}$ . In air, the index of refraction is very close to 1, so the speed of light in air is considered to be equal to that of vacuum. Other transparent media have indices of refraction greater than 1.

As a ray of light crosses the boundary between two media, its frequency remains unchanged (since frequency depends only on the source of the light), but its speed of propagation, and therefore wavelength, changes. This results in a change of direction of the ray. The direction is determined by the angle the ray makes with the normal of the boundary between the media. The law relating the angle of incidence $\theta_{i}$ to the angle of refraction $\theta _{r}$ is called Snell's Law: $\frac{\sin \theta _{i}}{\sin \theta _{r}}=\frac{n_{r}}{n_{i}}$ . Alternatively, it can be written as $n_{i}\sin \theta _{i}=n_{r}\sin \theta _{r}$ .

Notice that if $\theta _{i}=0$ , that is, the incident light is perpendicular to the boundary, then $\theta _{r}=0$ as well, so the light will not be refracted.

From Snell's Law, it follows that when light crosses the boundary to the medium with the greater index of refraction (for example, from air to water), the refracted ray will be closer to the normal than the incident ray. Also, the expression for the angle of refraction $\sin \theta _{r}=\frac{n_{i}\sin \theta_{i}}{n_{r}}$ will always have solutions, since the right-hand side of the equation will be less than 1 for any incident angle. However, when light goes from a medium with a greater index of refraction to a medium with a smaller one (as with from water to air), the refracted ray will be further away from the normal than the incident ray.

It can make a 90° angle with the normal, but it cannot go further, since it would then longer be in the refractive medium and would reflect back to the incident medium. This phenomenon is known as total internal reflection. When the incident angle is larger than a certain value called the critical angle, light will not refract, just reflect. The value of the critical angle can be found by setting the angle of refraction to 90°, which means $\sin \theta _{r}=1$ . Then, $\sin \theta _{c}=\frac{n_{r}}{n_{i}}$ .

Review the physics of why lenses work in The Law of Refraction. Refraction and reflection usually come together; their interplay is especially visible in Total Internal Reflection.

7c. Explain the interference pattern in a double-slit experiment and what these results mean

In the double-slit experiment, light is emitted onto a sheet with two small openings, and a pattern is observed on a screen some distance away. If the light is modeled as a beam of particles, what pattern should be observed on the screen?
If light is instead modeled as a wave, what pattern should be observed? What wave property explains this pattern?

The double-slit experiment conclusively demonstrated the wave nature of light. If light was a beam of particles, the pattern on the screen in the double-slit experiment would have consisted of two bright spots in front of the slits. Instead, the double-slit experiment yielded a pattern of alternating bright and dark bands. We can explain this pattern, and predict the location of the bright and dark bands, by explaining light as a wave.

Two light waves with the same frequency and same initial phase leave the slits. When the waves reach the screen, they interfere with one another. By the time they reach the screen, they have traveled different distances and are no longer in phase. The difference between these distances each wave travels is called path difference and is denoted by $\delta$ .

If the path difference equals an integer number $m$ of wavelengths: $\delta=m \lambda$ , then the waves undergo constructive interference, and their amplitudes add up. In this case, we observe a bright band. However, if the path difference is an odd multiple of half wavelengths: $\delta =(m+\frac{1}{2})\lambda$ , the waves will undergo destructive interference and cancel each other out. In this case, we observe a dark band.

From geometric considerations and the assumption that the screen is far away from the sheet with the slits, the locations of bright and dark bands can be found as $y_{bright}=\frac{m\lambda L}{a}$ and $y_{dark}=\frac{(m+\frac{1}{2})\lambda L}{a}$ . Here, $L$ is the distance between the sheet and the screen and a is the distance between the slits. Note that the zeroth bright band occurs at $y=0$ , in the center of the screen, opposite the midpoint between the slits. Note also that the dark and bright bands will be equidistant from one another as long as they are close enough to the center of the screen. Further away from the center of the screen, the brightness of the bright bands becomes less intense, and the bright bands spread further away from one another.

Review Figure 27.13 in Young's Double Slit Experiment.

7d. Explain how rainbows are produced

Why do light waves of different frequencies refract at different angles?
How do these frequencies and angles correspond to the different colors of visible light?

Snell's Law describes how the angle of refraction depends on the ratio of the indices of refraction of the incident and refractive media. For many media (including water and glass), the index of refraction depends slightly on the frequency or wavelength of incident light. This dependency is called dispersion, and a medium for which $n$ depends on $f$ (or $\lambda$ ) is called dispersive.

A rainbow is the most familiar natural phenomenon that demonstrates dispersion. The light from the sun contains waves with all frequencies of the visible spectrum. The combination of all these waves is perceived as white light. If there are water droplets in the air, the light refracts when it enters the droplets, and then refracts again as it leaves the droplets. Since the angle of refraction is different for light of different frequencies, waves of different colors separate. Observers see this as a rainbow.

Review how the dependence of refractive index on the color of light can cause light rays of different colors to take different paths in Dispersion: The Rainbow and Prisms.

7e. Explain how the Huygens principle leads to diffraction at a single slit

Draw a sketch of the light rays sent out by a star. Then add lines representing the wavefronts to this drawing. How is this similar to the ripples in a pond when a pebble is thrown in?
Which general principle of wave physics makes it possible to construct a straight wavefront from many circular wavefronts?
Why does diffraction prove that light is a wave and not a stream of particles?

As Maxwell predicted, you do not need to know that light is a wave to understand most optical instruments. It is enough to describe light by rays that follow the geometric laws of refraction and reflection. This is because the wavelength of visible light is much shorter than the wavelength of more familiar waves, such as water waves. But the double-slit experiment shows that light is a wave since it forms an interference pattern between two overlapping waves coming from different slits. Diffraction answers the question how these two waves were able to overlap in the first place, after they came through the slits: they can go around corners.

To explain diffraction, think of every point in a slit as the source of a spherical wave, also called a wavelet. The wavefronts coming from this point source look like concentric circles, whereas the rays for the same point source form a star-shaped pattern. This means that rays and wavefronts intersect each other at right angles. When we put many of these point sources next to each other, their wavefronts combine to form a single, new, wavefront. The superposition principle that applies to all harmonic waves makes this possible.

Huygens' principle uses this idea to explain how wavefronts propagate, even around obstacles. It does not involve drawing any rays at all, only wavefronts. The principle states: Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets.

You can use this to construct the wavefronts of a light wave with straight wavefronts passing through a slit. The result is that the wavelets at the corners are tangent to a new, curved, wavefront. This curved wavefront propagates outward from the slit in a fan shape, so that light coming from two neighboring slits can overlap and interfere. When a single slit gets very narrow, Huygens' principle says it will act more and more like a single point source of light, with wavefronts that become almost semicircles behind the slit, allowing light to spread out in all directions.

For a single slit that is not too narrow, all of the points inside the slit create their individual wavelets, and these may have path-length differences between them that lead to destructive interference in certain directions, just as we have discussed for the two-slit interference. This means that even a single slit will show an interference pattern if one looks closely at the light reaching a screen behind the slit.

These ideas help explain most of the wavelength-dependent effects you can observe with light. Review Huygens' Principle: Diffraction and see an example of a single slit in Single Slit Diffraction.

7f. Use the Rayleigh criterion for the resolution limit at different wavelengths

Astronomers cannot use binoculars to see the moons of Jupiter. Why do they need to use a telescope with a large diameter instead?
Draw two dots next to each other on the wall, then back away from the wall. At some point, you will not be able to see the dots as distinct anymore. They will blur into one. Why does this happen?
Which two quantities can be changed to reduce the angle at which two light sources can still be resolved as two separate spots by an optical instrument?

For a single slit with the width $D$ , the light collected on a screen at angle $\theta$ relative to the forward direction will show destructive interference, leading to dark regions, at angles that satisfy the condition:

$D\sin \theta =m\lambda$ , for $m$ = 1, –1, 2, –2, 3,… .

Here, $\lambda$ is the wavelength of the light. The brightest direction behind the slit is the forward direction, corresponding to $\theta =0$ . But this central bright line (roughly in the shape of the slit itself) is surrounded by darkness at the angle of the first destructive interference, where $m=1$ or $-1$ . The width of the bright line can be estimated from this angle range. The narrower the slit is, the broader the bright region becomes, because diffraction is more pronounced.

We can do the same if we replace the slit with a circular hole. On a screen behind the hole, we will see a bright spot surrounded by a dark circle of destructive interference. The smaller the hole, the larger the bright spot will be. By calculating the diffraction and interference pattern for this hole, one can find the formula that describes the size of the bright spot. It is usually given in terms of the angle $\theta$ between the forward direction and a ray that goes from the hole to the edge of the bright spot. Aperture is the diameter $D$ of the hole. The angle characterizing the bright spot is given by:

$\theta = 1.22 \frac{\lambda}{D}$ .

The shorter the wavelength $\lambda$ , the smaller the bright spot will be. In this formula, it is assumed that angles are measured in radians, not in degrees. This is not important in the earlier formulas involving $\sin \theta$ , but you will get the wrong result if you do not use radians here.

This case of a circular aperture can be used to estimate how close together two small light sources can be brought, before the bright spots they create behind that same aperture overlap and merge into a single spot. The angle at which this happens is called the resolution limit. In terms of the angle between the light sources, the Rayleigh criterion says that the smallest allowable angle is just equal to the angular size of a single bright spot as given above: $\theta = 1.22\ \lambda/ D$ . This means that we must either use a large aperture $D$ or a short wavelength $\lambda$ to resolve two light sources that are at a small angle to each other.

In astronomy, telescopes with a large aperture are always better. The reason is that astronomers try to resolve celestial objects that may not only be far away (and therefore may have low intensity) but also close together.

For example, to see the moons orbiting a planet such as Jupiter, or to see Saturn's rings, we have to think of each of these as separate sources that send their light through the aperture of our telescope. The angle between Jupiter and its moon Europa, as seen from our telescope, is very small. So we want a large aperture so our telescope can create two clear, distinct bright spots for Jupiter and Europa on the detector, camera film, or on the retina of the eye that looks through that aperture from the other end.

If you cannot change the aperture or wavelength, you can still affect the resolution your eye can achieve. Just move closer to the object you want to resolve. This can also be explained by the Rayleigh criterion.

The reason is that the distance between you and the object determines the angle at which different points of the object will appear from your point of view. The closer you are, the larger the angles get. In this case, the aperture is the diameter of your eye's pupil, and it stays roughly the same. So the resolution limit is the same, but by getting closer you make the object's angular size larger than the resolution limit.

Review the drawing that illustrates how we measure angles with respect to the aperture in Limits of Resolution: The Rayleigh Criterion.

7g. Explain how polarized light is created and detected

When you take two pairs of 3D glasses in a movie theater and hold them against the light, you can change the brightness by looking through both combined and rotating one of them. Why?
Why do polarized sunglasses reduce glare from water surfaces?
When you look at a laptop screen with polarized sunglasses, you may see the brightness change if you tilt your head. Why?

Polarized light is created when the electromagnetic wave that the light is made of shows a particularly regular oscillation. This manifests itself in the fact that the electric field at a given point in the wave oscillates back and forth only along a single line: for example up and down, or left and right. The former would be called vertical polarization, the latter horizontal polarization. We learned earlier that the electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and also perpendicular to the direction in which the wave is traveling.

For light created by most sources, such as stars or light bulbs, the light waves that propagate in any given direction are not polarized, because they consist of wave trains (or wave packets) that randomly oscillate with the electric fields pointing in arbitrary directions perpendicular to the general forward direction. But one can create polarized light by sending such unpolarized light waves through a polarization filter.

A special type of adjustable polarization filter can be made using liquid crystals. They change their orientation in response to electric fields, and this orientation determines whether they let certain polarizations pass through. This effect is used in computer and phone displays. It is described by the following formula, which assumes two polarization filters whose polarization axes are tilted relative to each other by an angle $\theta$ . Then the transmitted light intensity $I$ is related to the intensity $I_{o}$ of the original light beam by:

$I=I_{o}\cos^{2} \theta$ .

The square of the cosine means that you get perfect transmission ( $I=I_{o}$ ), for two angles: $\theta$ equal to 0° and $\theta$ equal to 180°.

Another way to create polarized light is by reflecting unpolarized light from a semi-transparent surface, such as water. The amount of light that is reflected by the surface, compared to the amount transmitted into the material, depends on the polarization of the electromagnetic wave. In particular, if the electric field is in the same plane as the lines that follow the propagation directions of the incident and reflected light, there is a special angle of incidence at which no light is reflected at all. This is called the Brewster angle, $\theta_{b}$ . It depends on the index of refraction of the materials.

For example, if the light comes in from the air and hits a water surface, the Brewster angle is given by $\theta _{b}=\arctan(1/n)$ , where $n=1.33$ is the index of refraction for water.

Review Polarization for examples of polarization and how to create it.

Unit 7 Vocabulary

You should be familiar with these terms to complete the final exam.

aperture
Brewster angle
concave mirror
converging lens
convex mirror
diffraction
dispersion
diverging lens
focal length
focal point
image
index of refraction
interference (constructive and destructive)
lens equation
magnification
mirror equation
object
path difference
plane mirror
polarization
polarization filter
Rayleigh criterion
real Image
reflection (specular and diffuse)
refraction
resolution limit
Snell's Law
total internal reflection
virtual Image