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Optics

Written by tutor Andrea L.

In this section we examine several wavelike behaviors of light, specifically: reflection, refraction, diffraction and interference. Visible light is a type of electromagnetic wave, whose wavelength ranges between about 7x10-7 meters (red light) and 4x10-7 meters (violet light). Like with other types of waves, the speed (v), frequency (f), and wavelength (λ) of light are related by: v = fλ. Unless otherwise noted, we may take the speed of light to be equal to its speed in a vacuum, c (c = 3x108 m/s).

Reflection and Refraction

When examining the behavior of light at a boundary surface, there are three types of rays we often discuss:

1) The original incident ray, traveling at angle θ1
2) The part of the incident ray which reflects from the surface (the reflected ray, which bounces off the surface at angle θ1’)
3) The part of the incident ray that travels into the surface (the refracted ray, traveling at angle θ2)

The angles of these three rays are always measured with respect to a line called the normal that is perpendicular to the surface at the point where the incident ray encounters the surface.

Reflection

According to the Law of Reflection, θ1 = θ1’. The angle of incidence is always equal to the angle of reflection, except on the opposite side of the “normal.”

Refraction

To determine the angle of refraction, we need to consider both the angle of incidence and the speed of light in the two media. Although the frequency of the light remains the same in both media, the wavelength and velocity may be different, causing the light ray to bend towards or away from the normal. The speed of light in a medium is described by the index of refraction, n, which is defined as n = c/v (c = speed of light in a vacuum, 3x108 m/s; v = speed of light in the medium). Note that since v must be less than c, n is always ≥ 1.

This information may be used to calculate the angle of refraction via Snell’s Law:

n1sin(θ1) = n2sin(θ2

where n1 and n2 are the indices of refraction of the two media. Depending on the incident angle and the indices of refraction, we may observe different behavior in the refracted ray:

-If θ1 = 0° (the incident ray is exactly along the normal), θ2 is also 0°, regardless of n1 and n2.
-For all other incident angles, there are three possible cases:
          1) If n2 = n1, then θ2 = θ1, and the light ray continues in its original direction.

          2) If n2 > n1, then θ2 <>1. The ray slows down when it enters the second medium, bending towards the normal.

          3) If n2 <>1, then θ2 > θ1. The ray speeds up when it enters the second medium, bending away from the normal.

Total internal reflection

When a light ray encounters a boundary surface such that n2 <>1, there is a particular angle of incidence for which all the light refracts at exactly 90° (parallel to the surface). The angle of incidence in this situation is called the critical angle, θc, and may be calculated with: θc = sin-1(n2/n1). At this angle and any larger angle of incidence, light cannot escape the first medium. When θ1 > θc, all the light is reflected so there is no refracted ray. This behavior is called total internal reflection.

Chromatic dispersion

Here we briefly consider how the angle of refraction depends on the color of incoming light. Different colors of light have different wavelengths, and the index of refraction n varies depending on the wavelength of the light. Generally, n is greater for shorter wavelengths (violet/blue light, for example) and smaller for longer wavelengths (red light). This means that violet light will be bent more than red light. Since white light is made up of all different colors, this property causes chromatic dispersion of incoming white light – the different colors are separated when the light enters a different medium, because of their different refraction angles. Note that we assume monochromatic (single- wavelength) light in all other sections.

Reflecting and refracting surfaces

There are specific reflecting and refracting surfaces which we will examine in detail. The reflecting surfaces fall into three categories: plane mirrors, concave mirrors, and convex mirrors. The refracting surfaces fall into two categories: convex lenses and concave lenses. As we will see, there are many similarities between concave mirrors and convex lenses, and between convex mirrors and concave lenses. In all of the following cases, “in front” of the mirror or lens will mean on the same side as the incident light and “behind” will mean on the opposite side from the incident light.

Mirrors

An object placed in front of a mirror can be thought of as a source of light. We wish to determine the location and properties of the object’s image, as created by the mirror. The light from this object spreads out as it travels towards the mirror, so is incident on the mirror at different angles. We can locate the image created by this object by drawing a few representative incident rays and their corresponding reflections (making use of the Law of Reflection, described above). The top of the image occurs where the reflected rays intersect, and the distance of this point from the mirror is called d i while the original object distance is called do. Images may be either real or virtual, depending on the type of mirror and object distance (see specifics below). Real images can be formed onto a surface, such as film or a screen, while virtual images exist only because in certain cases our brain perceives light rays as having come from a point where rays do not actually converge.

A couple important points to keep in mind for mirrors:

-Real images always form on the same side of a mirror as the object and are inverted, and virtual images always form on the opposite side and are the same orientation as the object
-Image distance is positive when the image is real and negative when the image is virtual

Plane mirrors

To see how virtual images work, we can examine the case of a plane mirror. As we see in this case, the reflected rays only intersect if we extend them backwards (behind the mirror).

Since there are not actually any light rays at the point where the image is created, this is called a virtual image. We find that the image in this case is created at the same distance behind the mirror as the object is in front of it. By convention, object distances are taken to be positive, and image distances for virtual images are taken to be negative. So in the plane mirror case, di = -do.

Spherical mirrors

Spherical mirrors are mirrors in the shape of a small section of the surface of a sphere and occur in two types: concave and convex. Before we can locate images created by spherical mirrors, there are a few important quantities we need to define:

-The center of curvature, C, of a spherical mirror is the distance from the mirror to the center of the sphere of which the mirror is a part.
-The central axis is a line that extends through the center of curvature and through the center of the mirror.
-The focal point of a mirror is the point at which rays incident from infinitely far away and parallel to the central axis would converge upon reflection.
          o For concave mirrors, the reflected rays really do converge, at a point in front of the mirror, so this
          is a “real” focal point.
          o For convex mirrors, the parallel incident rays are not really reflected through a common point;
          however, the eye interprets the reflections as having come from a single point behind the mirror. This perceived
          point is a “virtual” focal point.
          o The focal length is defined as being positive for a concave mirror and negative for a convex mirror
          o For both concave and convex mirrors, the distance to the center of curvature and distance to the
          focal point are related by: f = ½ C

Concave mirrors

To locate images created by spherical mirrors, there are specific incident and reflected rays that are useful to draw. Which rays we choose to draw depends somewhat on the position of the object with relation to the focal point and center of curvature. We could draw any two of these four special rays:

1. A ray initially parallel to the principal axis, that then reflects through the focal point
2. A ray that passes through the focal point, then reflects parallel to the central axis
3. A ray that passes through the center of curvature, then reflects along itself
4. A ray incident towards the center of the mirror, that then reflects symmetrically about the principal axis

The point where the rays intersect indicates the location at which the image appears. Five specific cases may be considered (in the following figures, note that black arrow represent objects, light blue arrows represent images, and the numbers next to the rays denote which of the “special rays” is drawn)






As we can see, as the object is moved closer to the mirror, the image gets larger and moves further from the mirror. If the object is at the focal point, no image appears. When the object moves in closer than the focal point, the image becomes virtual and on the opposite side of the mirror.

Convex mirrors

To locate images created by convex mirrors, we can draw just two rays:

1. A ray parallel to the principal axis, that then reflects as if it came from the virtual focal point
2. A ray incident towards the virtual focal point, then reflected parallel to the principal axis

For convex mirrors, we need only consider one case:

For both concave and convex mirrors, the focal point (f), image distance (di), or object distance (do) can be calculated exactly from the following equation, if the other two quantities are known:

1/f = 1/di + 1/do

We may also calculate a mirror’s magnification (M). This describes the ratio of image height (h i) to object height (ho) and is given by:

M = di/do = hi/ho

Note that a negative magnification or object height simply means the image is inverted compared to the original object.

Refracting Surfaces

There are two types of refracting surfaces we will consider: convex lenses and concave lenses. In both cases, we assume that the light begins to refract at a line drawn vertically through the center of the lens (this is called the thin lens approximation).

A couple important points to keep in mind for lenses:

-Real images will form on the side of the lens opposite the object, and virtual images form on the same side of the lens as the object. (this is the opposite to the behavior for mirrors)
-The focal length f is defined as being positive for convex lenses and negative for concave lenses (also opposite to the convention for mirrors)
-Image distance is positive when the image is real and negative when the image is virtual (same as for mirrors)

Like with mirrors, we can again relate focal length, the object and image distances, and magnification with the equations:

1/f = 1/di + 1/do and M = di/do = hi/ho

Convex (“converging”) lenses

Convex lenses have two real focal points, one on either side of the lens. Let us define f1 as the focal point in front of the lens, and f2 as the focal point behind the lens. We can locate the image created by drawing a ray diagram using the following rays:

1. A ray parallel to the principal axis, which is refracted through f2
2. A ray through f1, which is refracted parallel to the principal axis
3. A ray incident towards the center of the lens, which continues through in the same direction

Note that if the object is inside the focal point, we need to extend each of these rays back towards the object’s side of the lens to see where they intersect to form an image.

Like with concave mirrors, there are five separate cases we can consider for convex lenses:






Concave (“diverging”) lenses

Concave lenses have two virtual focal points, defined f1 and f2 as above. In this case, we draw:

1. A ray parallel to the principal axis, which diverges away from f1
2. A ray aimed towards f2, which emerges parallel to the axis
3. A ray incident towards the center of the lens, which continues through in the same direction

For concave lenses, like with convex mirrors, we need only consider one case:

Two-lens systems

There are many applications which utilize multiple thin lenses; compound microscopes and refracting telescopes are two common examples. The final image created by a two-lens system can be found by splitting the problem into two single-lens problems:

1. Ignoring the presence of lens 2, locate the image distance di of the image produced by lens 1.
2. Now ignoring the presence 1, treat the image from step 1 as the object for lens 2 and find the image distance for this new object.

For multiple-lens systems, the total magnification is simply the product of the magnifications of the individual lenses.

Diffraction and Interference

When a light wave encounters a barrier with an opening of similar size to the wavelength of the light, it “diffracts” or spreads out in a semi-circular pattern on the opposite side of the barrier. The narrower the slit, the more the wave spreads out on the opposite side. When two waves come from nearby points, such as slits in a screen, they will overlap and “interfere.” Evidence of this overlapping is observed if a viewing screen is placed some distance away from the slits, since we see an interference pattern of alternating bright and dark bands (sometimes termed bright and dark “fringes”).

-Bright bands mean:
          o Constructive interference (the maxima of the waves line up)
          o The interfering beams have traveled along paths with a length difference (ΔL) of 0, or equal to an
          integer number of wavelengths of the light (ΔL = 0, λ, 2λ, ...)
-Dark bands mean:
          o Destructive interference (one wave’s maximum lines up with the other’s minimum)
          o The interfering beams have traveled along paths with a length difference equal to a half- integer
          number of wavelengths (ΔL = ½ λ, 3/2 λ, ...)

Double-slit diffraction

In double-slit diffraction, each slit acts as a point source of light, from which semi-circular wave fronts originate and interfere. In this setup, the central axis is defined as a line straight from a point halfway between the slits to the screen where the interference pattern is observed, and the locations of bright and dark fringes on the screen can be identified by their angles θ from this central axis. If the slits are a distance d apart, the path length difference (ΔL) between the waves from the two slits is given by: ΔL = d sin(θ)

-Bright fringes are observed at points on the screen where ΔL is an integer number of wavelengths, so:
          ΔL = mλ
          mλ = d sin(θ)
-For dark fringes, ΔL must be a half-integer number of wavelengths, so:
          ΔL = (m + ½) λ
          (m + ½) λ = d sin(θ)

In these equations, m is an integer meaning the number of bright or dark fringes away from the central axis. Sometimes m is called the “order” of the maximum or minimum, and it also tells by how many wavelengths the paths of the interfering waves differ. The central maximum is called the “0 order maximum” because there is no difference in the paths traveled by the two waves at this point.

After calculating the angle that corresponds to a particular maximum or minimum, its distance y along the screen from the central axis can be determined with the tangent function: tan(θ) = y/D

Single-Slit Diffraction

The methods used in this case are similar for double slit diffraction, though we can only calculate the locations of the dark fringes due to more involved mathematics. Since there is just one slit in this case, we look at the path length difference between rays coming from two specific points along the slit, rather than from two separate slits. In order for there to be a dark fringe at a certain point on the screen, each ray we could choose must be paired with a ray whose path length is different by (m + ½)λ at that point. Rather than examining each possible pair of rays here, we will simply quote the result. It is found that dark fringes are located at points along the screen where the path difference, ΔL, for rays originating at the top and bottom of the slit is an integer number of wavelengths. Although the top and bottom rays interfere constructively in this case, each experiences destructive interference with a ray from another part of the slit. So we can locate the dark fringes with the equation:

mλ = a sin(θ)

Note that for a narrower slit width a, the angle θ of the dark fringes must be greater, so the pattern will become more spread out on the screen. Again m represents the “order,” how many fringes we are away from the central axis.

Diffraction gratings

This setup is again similar to double slit diffraction, only now we have a much greater number of slits, N. Similarly for diffraction gratings, the path length difference between adjacent rays is again ΔL = d sin(θ). If the distance d between adjacent slits is not given directly, we can calculate it with d = w/N, if N slits occupy a total width w. Like in the double slit case, bright lines will be located at angles from the central axis given by mλ = d sin(θ).

Optics Quiz

An object is placed a distance 1.5f from a converging lens of focal length f. What is the location of the image?

A. 1.5f, on the opposite side of the lens
B. 3f, on the opposite side of the lens
C. 0.5f, on the same side of the lens as the object
D. 3f, on the same side of the lens as the object
The correct answer here would be B.

An object is placed a distance of 6cm from a concave mirror, and a real image is observed at a distance of 9cm. What is the radius of curvature of this mirror?

A. 3cm
B. 3.6cm
C. 7.2cm
D. 15cm
The correct answer here would be C.

A blue light (λ = 4.75x10-7 m) shines through a double-slit screen, creating an interference pattern. How will this pattern change when a red light (λ = 6.50x10-7 m) is shone through the same screen instead?

A. The pattern will become more spread out
B. The pattern will become more compressed
C. The fringes will remain at the same locations, but become brighter
D. The fringes will remain at the same locations, but become dimmer
The correct answer here would be A.

A monochromatic light ray passes from air (n = 1.00) into water (n = 1.33). The ray is incident upon the surface at an angle of 30° with respect to the normal. Which is most nearly the angle of refraction?

A. 17°
B. 22°
C. 30°
D. 42°
The correct answer here would be B.

Which of the following best describes the image formed by an object placed at a distance of 5cm from a convex lens of focal length f = 10cm?

A. Virtual, upright, larger than object
B. Real, inverted, larger than object
C. Real, inverted, smaller than object
D. None of the above; no image is formed
The correct answer here would be A.

A yellow light (λ = 5.7x10-7 m) shines on a single slit of width 2.5x10-6 m. At what angle from the central axis will the second dark fringes be observed?

A. 13°
B. 21°
C. 27°
D. 35°
The correct answer here would be A.

A light ray traveling through an unknown material encounters a boundary with a glass surface (n2 = 1.50). For small incident angles, a refracted ray is observed in the glass. However, at any incident angle larger than 58°, the ray is entirely reflected back into the first material. Approximately what is the index of refraction of this unknown material?

A. 1.30
B. 1.77
C. 2.10
D. 2.83
The correct answer here would be B.

At a particular point P on a viewing screen, the path length difference ΔL between beams from two narrow slits is 5/2 λ. What is observed at point P?

A. A fifth-order dark fringe
B. A fifth-order bright fringe
C. A third-order dark fringe
D. A second-order dark fringe
The correct answer here would be D.

Which of the following is not true about refraction?

A. If a ray is incident along the normal when it encounters a boundary surface, its direction of motion will remain the same as it enters the second material, regardless of the indices of refraction of the two materials
B. A ray passing from water (n = 1.33) into air (n = 1.00) will refract away from the normal
C. All wavelengths of light must refract at the same angle when entering a different medium
D. A ray passing from air (n = 1.00) into glass (n = 1.50) could never experience total internal reflection, for any incident angle
The correct answer here would be C.

Green light (λ = 5.1x10-7 m) is incident upon a diffraction grating of width 3cm. The first-order maxima are located at 10° from the central axis. Approximately how many slits does this diffraction grating have?

A. 1135
B. 3404
C. 10,214
D. 20,118
The correct answer here would be C.
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