Diffraction occurs when light passes through a single slit. Rank the following three choices in decreasing order, according to the extend of the diffraction that occurs (largest diffraction first) Blue light, narrow slit Red light, narrow slit. Bending of light near the edges of an obstacle or slit and spreading into the region of geometrical shadow is known as diffraction of light. The diffraction phenomenon is classified into two types: 1. Fraunhofer diffraction: The source of light and the screen on which the diffraction pattern is obtained are effectively at infinite distance from the diffracting system. Tum efficiency (EQE) of a QW-LED depends on its injection efficiency, internal quantum efficiency (IQE), and light extraction efficiency (LEE). The IQE is defined as the ratio of radiative recombination rate to the sum of radiative and non-radiative recombination rates of QW-LEDs. By assuming the non-radiative recombination rate to be negligible at.
- On Which Factors Diffraction Of Light Depends On Light
- What Affects Diffraction
- On Which Factors Diffraction Of Light Depends On Water
The reflection and refraction of light
Rays and wave fronts
Light is a very complex phenomenon, but in many situations its behavior can be understood with a simple model based on rays and wave fronts. A ray is a thin beam of light that travels in a straight line. A wave front is the line (not necessarily straight) or surface connecting all the light that left a source at the same time. For a source like the Sun, rays radiate out in all directions; the wave fronts are spheres centered on the Sun. If the source is a long way away, the wave fronts can be treated as parallel lines.
Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.
The law of reflection
Objects can be seen by the light they emit, or, more often, by the light they reflect. Reflected light obeys the law of reflection, that the angle of reflection equals the angle of incidence.
For objects such as mirrors, with surfaces so smooth that any hills or valleys on the surface are smaller than the wavelength of light, the law of reflection applies on a large scale. All the light travelling in one direction and reflecting from the mirror is reflected in one direction; reflection from such objects is known as specular reflection.
Most objects exhibit diffuse reflection, with light being reflected in all directions. All objects obey the law of reflection on a microscopic level, but if the irregularities on the surface of an object are larger than the wavelength of light, which is usually the case, the light reflects off in all directions.
A plane mirror is simply a mirror with a flat surface; all of us use plane mirrors every day, so we've got plenty of experience with them. Images produced by plane mirrors have a number of properties, including:
- the image produced is upright
- the image is the same size as the object (i.e., the magnification is m = 1)
- the image is the same distance from the mirror as the object appears to be (i.e., the image distance = the object distance)
- the image is a virtual image, as opposed to a real image, because the light rays do not actually pass through the image. This also implies that an image could not be focused on a screen placed at the location where the image is.
A little geometry
Dealing with light in terms of rays is known as geometrical optics, for good reason: there is a lot of geometry involved. It's relatively straight-forward geometry, all based on similar triangles, but we should review that for a plane mirror.
Consider an object placed a certain distance in front of a mirror, as shown in the diagram. To figure out where the image of this object is located, a ray diagram can be used. In a ray diagram, rays of light are drawn from the object to the mirror, along with the rays that reflect off the mirror. The image will be found where the reflected rays intersect. Note that the reflected rays obey the law of reflection. What you notice is that the reflected rays diverge from the mirror; they must be extended back to find the place where they intersect, and that's where the image is.
Analyzing this a little further, it's easy to see that the height of the image is the same as the height of the object. Using the similar triangles ABC and EDC, it can also be seen that the distance from the object to the mirror is the same as the distance from the image to the mirror.
Light reflecting off a flat mirror is one thing, but what happens when light reflects off a curved surface? We'll take a look at what happens when light reflects from a spherical mirror, because it turns out that, using reasonable approximations, this analysis is fairly straight-forward. The image you see is located either where the reflected light converges, or where the reflected light appears to diverge from.
A spherical mirror is simply a piece cut out of a reflective sphere. It has a center of curvature, C, which corresponds to the center of the sphere it was cut from; a radius of curvature, R, which corresponds to the radius of the sphere; and a focal point (the point where parallel light rays are focused to) which is located half the distance from the mirror to the center of curvature. The focal length, f, is therefore:
focal length of a spherical mirror : f = R / 2
This is actually an approximation. Parabolic mirrors are really the only mirrors that focus parallel rays to a single focal point, but as long as the rays don't get too far from the principal axis then the equation above applies for spherical mirrors. The diagram shows the principal axis, focal point (F), and center of curvature for both a concave and convex spherical mirror.
Spherical mirrors are either concave (converging) mirrors or convex (diverging) mirrors, depending on which side of the spherical surface is reflective. If the inside surface is reflective, the mirror is concave; if the outside is reflective, it's a convex mirror. Concave mirrors can form either real or virtual images, depending on where the object is relative to the focal point. A convex mirror can only form virtual images. A real image is an image that the light rays from the object actually pass through; a virtual image is formed because the light rays can be extended back to meet at the image position, but they don't actually go through the image position.
To determine where the image is, it is very helpful to draw a ray diagram. The image will be located at the place where the rays intersect. You could just draw random rays from the object to the mirror and follow the reflected rays, but there are three rays in particular that are very easy to draw.
Only two rays are necessary to locate the image on a ray diagram, but it's useful to add the third as a check. The first is the parallel ray; it is drawn from the tip of the object parallel to the principal axis. It then reflects off the mirror and either passes through the focal point, or can be extended back to pass through the focal point.
The second ray is the chief ray. This is drawn from the tip of the object to the mirror through the center of curvature. This ray will hit the mirror at a 90° angle, reflecting back the way it came. The chief and parallel rays meet at the tip of the image.
The third ray, the focal ray, is a mirror image of the parallel ray. The focal ray is drawn from the tip of the object through (or towards) the focal point, reflecting off the mirror parallel to the principal axis. All three rays should meet at the same point.
A ray diagram for a concave mirror is shown above. This shows a few different things. For this object, located beyond the center of curvature from the mirror, the image lies between the focal point (F) and the center of curvature. The image is inverted compared to the object, and it is also a real image, because the light rays actually pass through the point where the image is located.
With a concave mirror, any object beyond C will always have an image that is real, inverted compared to the object, and between F and C. You can always trade the object and image places (that just reverses all the arrows on the ray diagram), so any object placed between F and C will have an image that is real, inverted, and beyond C. What happens when the object is between F and the mirror? You should draw the ray diagram to convince yourself that the image will be behind the mirror, making it a virtual image, and it will be upright compared to the object.
A ray diagram for a convex mirror
What happens with a convex mirror? In this case the ray diagram looks like this:
As the ray diagram shows, the image for a convex mirror is virtual, and upright compared to the object. A convex mirror is the kind of mirror used for security in stores, and is also the kind of mirror used on the passenger side of many cars ('Objects in mirror are closer than they appear.'). A convex mirror will reflect a set of parallel rays in all directions; conversely, it will also take light from all directions and reflect it in one direction, which is exactly how it's used in stores and cars.
The mirror equation
Drawing a ray diagram is a great way to get a rough idea of how big the image of an object is, and where the image is located. We can also calculate these things precisely, using something known as the mirror equation. The textbook does a nice job of deriving this equation in section 25.6, using the geometry of similar triangles.
In most cases the height of the image differs from the height of the object, meaning that the mirror has done some magnifying (or reducing). The magnification, m, is defined as the ratio of the image height to the object height, which is closely related to the ratio of the image distance to the object distance:
A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.
What does a positive or negative image height or image distance mean? To figure out what the signs mean, take the side of the mirror where the object is to be the positive side. Any distances measured on that side are positive. Distances measured on the other side are negative.
f, the focal length, is positive for a concave mirror, and negative for a convex mirror.
When the image distance is positive, the image is on the same side of the mirror as the object, and it is real and inverted. When the image distance is negative, the image is behind the mirror, so the image is virtual and upright.
A negative m means that the image is inverted. Positive means an upright image. Eclipse for java ee.
Steps for analyzing mirror problems
On Which Factors Diffraction Of Light Depends On Light
There are basically three steps to follow to analyze any mirror problem, which generally means determining where the image of an object is located, and determining what kind of image it is (real or virtual, upright or inverted).
- Step 1 - Draw a ray diagram. The more careful you are in constructing this, the better idea you'll have of where the image is.
- Step 2 - Apply the mirror equation to determine the image distance. (Or to find the object distance, or the focal length, depending on what is given.)
- Step 3 - Make sure steps 1 and 2 are consistent with each other.
A Star Wars action figure, 8.0 cm tall, is placed 23.0 cm in front of a concave mirror with a focal length of 10.0 cm. Where is the image? How tall is the image? What are the characteristics of the image?
The first step is to draw the ray diagram, which should tell you that the image is real, inverted, smaller than the object, and between the focal point and the center of curvature. The location of the image can be found from the mirror equation:
which can be rearranged to:
The image distance is positive, meaning that it is on the same side of the mirror as the object. This agrees with the ray diagram. Note that we don't need to worry about converting distances to meters; just make sure everything has the same units, and whatever unit goes into the equation is what comes out.
Calculating the magnification gives:
Solving for the image height gives:
The negative sign for the magnification, and the image height, tells us that the image is inverted compared to the object.
To summarize, the image is real, inverted, 6.2 cm high, and 17.7 cm in front of the mirror.
Example 2 - a convex mirror
The same Star Wars action figure, 8.0 cm tall, is placed 6.0 cm in front of a convex mirror with a focal length of -12.0 cm. Where is the image in this case, and what are the image characteristics?
Again, the first step is to draw a ray diagram. This should tell you that the image is located behind the mirror; that it is an upright, virtual image; that it is a little smaller than the object; and that the image is between the mirror and the focal point.
The second step is to confirm all those observations. The mirror equation, rearranged as in the first example, gives:
Solving for the magnification gives:
This gives an image height of 0.667 x 8 = 5.3 cm.
All of these results are consistent with the conclusions drawn from the ray diagram. The image is 5.3 cm high, virtual, upright compared to the object, and 4.0 cm behind the mirror.
When we talk about the speed of light, we're usually talking about the speed of light in a vacuum, which is 3.00 x 108 m/s. When light travels through something else, such as glass, diamond, or plastic, it travels at a different speed. The speed of light in a given material is related to a quantity called the index of refraction, n, which is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
index of refraction : n = c / v
When light travels from one medium to another, the speed changes, as does the wavelength. The index of refraction can also be stated in terms of wavelength:
Although the speed changes and wavelength changes, the frequency of the light will be constant. The frequency, wavelength, and speed are related by:
The change in speed that occurs when light passes from one medium to another is responsible for the bending of light, or refraction, that takes place at an interface. If light is travelling from medium 1 into medium 2, and angles are measured from the normal to the interface, the angle of transmission of the light into the second medium is related to the angle of incidence by Snell's law :
Diffraction; thin-film interference
We discussed diffraction in PY105 when we talked about sound waves; diffraction is the bending of waves that occurs when a wave passes through a single narrow opening. The analysis of the resulting diffraction pattern from a single slit is similar to what we did for the double slit. With the double slit, each slit acted as an emitter of waves, and these waves interfered with each other. For the single slit, each part of the slit can be thought of as an emitter of waves, and all these waves interfere to produce the interference pattern we call the diffraction pattern.
After we do the analysis, we'll find that the equation that gives the angles at which fringes appear for a single slit is very similar to the one for the double slit, one obvious difference being that the slit width (W) is used in place of d, the distance between slits. A big difference between the single and double slits, however, is that the equation that gives the bright fringes for the double slit gives dark fringes for the single slit.
To see why this is, consider the diagram below, showing light going away from the slit in one particular direction.
In the diagram above, let's say that the light leaving the edge of the slit (ray 1) arrives at the screen half a wavelength out of phase with the light leaving the middle of the slit (ray 5). These two rays would interfere destructively, as would rays 2 and 6, 3 and 7, and 4 and 8. In other words, the light from one half of the opening cancels out the light from the other half. The rays are half a wavelength out of phase because of the extra path length traveled by one ray; in this case that extra distance is :
The factors of 2 cancel, leaving:
The argument can be extended to show that :
The bright fringes fall between the dark ones, with the central bright fringe being twice as wide, and considerably brighter, than the rest.
Diffraction effects with a double slit
Note that diffraction can be observed in a double-slit interference pattern. Essentially, this is because each slit emits a diffraction pattern, and the diffraction patterns interfere with each other. The shape of the diffraction pattern is determined by the width (W) of the slits, while the shape of the interference pattern is determined by d, the distance between the slits. If W is much larger than d, the pattern will be dominated by interference effects; if W and d are about the same size the two effects will contribute equally to the fringe pattern. Generally what you see is a fringe pattern that has missing interference fringes; these fall at places where dark fringes occur in the diffraction pattern.
We've talked about what happens when light encounters a single slit (diffraction) and what happens when light hits a double slit (interference); what happens when light encounters an entire array of identical, equally-spaced slits? Such an array is known as a diffraction grating. The name is a bit misleading, because the structure in the pattern observed is dominated by interference effects.
With a double slit, the interference pattern is made up of wide peaks where constructive interference takes place. As more slits are added, the peaks in the pattern become sharper and narrower. With a large number of slits, the peaks are very sharp. The positions of the peaks, which come from the constructive interference between light coming from each slit, are found at the same angles as the peaks for the double slit; only the sharpness is affected.
Why is the pattern much sharper? In the double slit, between each peak of constructive interference is a single location where destructive interference takes place. Between the central peak (m = 0) and the next one (m = 1), there is a place where one wave travels 1/2 a wavelength further than the other, and that's where destructive interference takes place. For three slits, however, there are two places where destructive interference takes place. One is located at the point where the path lengths differ by 1/3 of a wavelength, while the other is at the place where the path lengths differ by 2/3 of a wavelength. For 4 slits, there are three places, for 5 slits there are four places, etc. Completely constructive interference, however, takes place only when the path lengths differ by an integral number of wavelengths. For a diffraction grating, then, with a large number of slits, the pattern is sharp because of all the destructive interference taking place between the bright peaks where constructive interference takes place.
Diffraction gratings, like prisms, disperse white light into individual colors. If the grating spacing (d, the distance between slits) is known and careful measurements are made of the angles at which light of a particular color occurs in the interference pattern, the wavelength of the light can be calculated.
Interference between light waves is the reason that thin films, such as soap bubbles, show colorful patterns. This is known as thin-film interference, because it is the interference of light waves reflecting off the top surface of a film with the waves reflecting from the bottom surface. To obtain a nice colored pattern, the thickness of the film has to be similar to the wavelength of light.
An important consideration in determining whether these waves interfere constructively or destructively is the fact that whenever light reflects off a surface of higher index of refraction, the wave is inverted. Peaks become troughs, and troughs become peaks. This is referred to as a 180° phase shift in the wave, but the easiest way to think of it is as an effective shift in the wave by half a wavelength.
Summarizing this, reflected waves experience a 180° phase shift (half a wavelength) when reflecting from a higher-n medium (n2 > n1), and no phase shift when reflecting from a medium of lower index of refraction (n2 < n1).
For completely constructive interference to occur, the two reflected waves must be shifted by an integer multiple of wavelengths relative to one another. This relative shift includes any phase shifts introduced by reflections off a higher-n medium, as well as the extra distance traveled by the wave that goes down and back through the film.
Note that one has to be very careful in dealing with the wavelength, because the wavelength depends on the index of refraction. Generally, in dealing with thin-film interference the key wavelength is the wavelength in the film itself. If the film has an index of refraction n, this wavelength is related to the wavelength in vacuum by:
A step-by step approach
Many people have trouble with thin-film interference problems. As usual, applying a systematic, step-by-step approach is best. The overall goal is to figure out the shift of the wave reflecting from one surface of the film relative to the wave that reflects off the other surface. Depending on the situation, this shift is set equal to the condition for constructive interference, or the condition for destructive interference.
Note that typical thin-film interference problems involve 'normally-incident' light. The light rays are not drawn perpendicular to the interfaces on the diagram to make it easy to distinguish between the incident and reflected rays. In the discussion below it is assumed that the incident and reflected rays are perpendicular to the interfaces.
A good method for analyzing a thin-film problem involves these steps:
Step 1. Write down , the shift for the wave reflecting off the top surface of the film.
Step 2. Write down , the shift for the wave reflecting off the film's bottom surface.
One contribution to this shift comes from the extra distance travelled. If the film thickness is t, this wave goes down and back through the film, so its path length is longer by 2t. The other contribution to this shift can be either 0 or , depending on what happens when it reflects (this reflection occurs at point b on the diagram).
Step 3. Calculate the relative shift by subtracting the individual shifts.
Step 4. Set the relative shift equal to the condition for constructive interference, or the condition for destructive interference, depending on the situation. If a certain film looks red in reflected light, for instance, that means we have constructive interference for red light. If the film is dark, the light must be interfering destructively.
Step 5. Rearrange the equation (if necessary) to get all factors of on one side.
Step 6. Remember that the wavelength in your equation is the wavelength in the film itself. Since the film is medium 2 in the diagram above, we can label it . The wavelength in the film is related to the wavelength in vacuum by:
Step 7. Solve. Your equation should give you a relationship between t, the film thickness, and either the wavelength in vacuum or the wavelength in the film.
Example - a film of oil on water
Working through an example is a good way to see how the step-by-step approach is applied. In this case, white light in air shines on an oil film that floats on water. When looking straight down at the film, the reflected light is red, with a wavelength of 636 nm. What is the minimum possible thickness of the film?
Step 1. Because oil has a higher index of refraction than air, the wave reflecting off the top surface of the film is shifted by half a wavelength.
Step 2. Because water has a lower index of refraction than oil, the wave reflecting off the bottom surface of the film does not have a half-wavelength shift, but it does travel the extra distance of 2t.
Step 3. The relative shift is thus:
Step 4. Now, is this constructive interference or destructive interference? Because the film looks red, there is constructive interference taking place for the red light.
Step 5. Moving all factors of the wavelength to the right side of the equation gives:
Note that this looks like an equation for destructive interference! It isn't, because we used the condition for constructive interference in step 4. It looks like a destructive interference equation only because one reflected wave experienced a shift.
Step 6. The wavelength in the equation above is the wavelength in the thin film. Writing the equation so this is obvious can be done in a couple of different ways:
Step 7. The equation can now be solved. In this situation, we are asked to find the minimum thickness of the film. This means choosing the minimum value of m, which in this case is m = 0. The question specified the wavelength of red light in vacuum, so:
What Affects Diffraction
This is not the only thickness that gives completely constructive interference for this wavelength. Others can be found by using m = 1, m = 2, etc. in the equation in step 6.
On Which Factors Diffraction Of Light Depends On Water
If 106 nm gives constructive interference for red light, what about the other colors? They are not completely cancelled out, because 106 nm is not the right thickness to give completely destructive interference for any wavelength in the visible spectrum. The other colors do not reflect as intensely as red light, so the film looks red.
Why is it the wavelength in the film itself that matters?
The light reflecting off the top surface of the film does not pass through the film at all, so how can it be the wavelength in the film that is important in thin-film interference? A diagram can help clarify this. The diagram looks a little complicated at first glance, but it really is straightforward once you understand what it shows.
Figure A shows a wave incident on a thin film. Each half wavelength has been numbered, so we can keep track of it. Note that the thickness of the film is exactly half the wavelength of the wave when it is in the film.
Figure B shows the situation two periods later, after two complete wavelengths have encountered the film. Part of the wave is reflected off the top surface of the film; note that this reflected wave is flipped by 180°, so peaks are now troughs and troughs are now peaks. This is because the wave is reflecting off a higher-n medium.
Another part of the wave reflects off the bottom surface of the film. This does not flip the wave, because the reflection is from a lower-n medium. When this wave re-emerges into the first medium, it destructively interferes with the wave that reflects off the top surface. This occurs because the film thickness is exactly half the wavelength of the wave in the film. Because a half wavelength fits in the film, the peaks of one reflected wave line up precisely with the troughs of the other (and vice versa), so the waves cancel. Destructive interference would also occur with the film thickness being equal to 1 wavelength of the wave in the film, or 1.5 wavelengths, 2 wavelengths, etc.
If the thickness was 1/4, 3/4, 5/4, etc. the wavelength in the film, constructive interference occurs. This is only true when one of the reflected waves experiences a half wavelength shift (because of the relative sizes of the refractive indices). If neither wave, or both waves, experiences a shift of , there would be constructive interference whenever the film thickness was 0.5, 1, 1.5, 2, etc. wavelengths, and destructive interference if the film was 1/4, 3/4, 5/4, etc. of the wavelength in the film.
One final philosophical note, to really make your head spin if it isn't already. In the diagram above we drew the two reflected waves and saw how they cancelled out. This means none of the wave energy is reflected back into the first medium. Where does it go? It must all be transmitted into the third medium (that's the whole point of a non-reflective coating, to transmit as much light as possible through a lens). So, even though we did the analysis by drawing the waves reflecting back, in some sense they really don't reflect back at all, because all the light ends up in medium 3.
Destructive interference is exploited in making non-reflective coatings for lenses. The coating material generally has an index of refraction less than that of glass, so both reflected waves have a shift. A film thickness of 1/4 the wavelength in the film results in destructive interference (this is derived below)
For non-reflective coatings in a case like this, where the index of refraction of the coating is between the other two indices of refraction, the minimum film thickness can be found by applying the step-by-step approach: