# Phase difference and path relationship questions

### Calculating phase difference - The Student Room

Conceptual Problems. 1 •. A phase difference due to path-length difference is observed for monochromatic visible light. Which phase . Determine the Concept The relationship between the slit separation d and the angular position θm of each. How would you work out phase difference between two points given the frequency, distance and phase difference = 2pi x path difference divided by lambda. For a complete wave, the wavelength varies in λ and the phase is changed through 2p. Let there be two waves with a path difference of λ. Then, the phase.

When these are spatially and colour filtered, and then split into two waves, they can be superimposed to generate interference fringes. A laser beam generally approximates much more closely to a monochromatic source, and it is much more straightforward to generate interference fringes using a laser.

The ease with which interference fringes can be observed with a laser beam can sometimes cause problems in that stray reflections may give spurious interference fringes which can result in errors. Normally, a single laser beam is used in interferometry, though interference has been observed using two independent lasers whose frequencies were sufficiently matched to satisfy the phase requirements.

White light interference in a soap bubble. The iridescence is due to thin-film interference. It is also possible to observe interference fringes using white light. A white light fringe pattern can be considered to be made up of a 'spectrum' of fringe patterns each of slightly different spacing. If all the fringe patterns are in phase in the centre, then the fringes will increase in size as the wavelength decreases and the summed intensity will show three to four fringes of varying colour.

Young describes this very elegantly in his discussion of two slit interference. Since white light fringes are obtained only when the two waves have travelled equal distances from the light source, they can be very useful in interferometry, as they allow the zero path difference fringe to be identified.

Traditionally, interferometers have been classified as either amplitude-division or wavefront-division systems. In an amplitude-division system, a beam splitter is used to divide the light into two beams travelling in different directions, which are then superimposed to produce the interference pattern.

### Wave Motion | - relation between phase difference and path-askIITians

The Michelson interferometer and the Mach-Zehnder interferometer are examples of amplitude-division systems. A laser is basically just a bunch of light waves that all have the same wavelength and are all lined up with one another.

Suppose you place a card in front of the laser beam with two slits in it, such that waves can only pass through two spots. You then measure the amount of light that hits the wall on the other side of the room at various points. Figure of laser beam passing through two slits towards opposite wall For the experiment to work, the slits have to be tiny compared to the distance from the card to the wall, but they have to be larger than a single wavelength of the light.

That means that if we choose a spot on the wall, two light waves will be hitting it; one from the top slit and one from the bottom slit.

## Phase Difference and Path Difference

As they get close to the wall, and close to one another, they will start to interfere. Figure of waves in phase passing through slits and becoming out of phase as they near the opposite wall above the top slit The light coming from the bottom slit has to come much further than the light from the top slit, so more wavelengths will be needed to travel the longer distance. The key is to compare the number of wavelengths it takes for each light wave to travel from the slit to the wall.

For constructive interference, the difference in wavelengths will be an integer number of whole wavelengths. For destructive interference it will be an integer number of whole wavelengths plus a half wavelength.

### Diffraction and constructive and destructive interference (article) | Khan Academy

Think of the point exactly between the two slits. The light waves will be traveling the same distance, so they will be traveling the same number of wavelengths. That means that there will always be constructive interference at that spot, so we will always see a bright spot on the wall in the middle.

At that point, one of the waves will hit the wall with a crest when the other hits with a trough, so they will effectively cancel one another out, resulting in a dark spot there. This will result in another bright spot on the wall. This pattern will keep alternating so that we get a pattern of light spots and dark spots, both above and below our center bright spot.

Figure of diffraction pattern on the opposite wall If your slits are further apart, the light waves will be coming from spots that are further apart. That means that their path lengths will be more different from one another, giving bright spots that are closer together.

We can pretend to divide our slit into pieces, and compare the path lengths of the light coming from these pieces to one another to discover what sort of interference pattern we will get when they interact.

## Write a relation between phase difference and path difference

They are an equal distance from the center of the slit, so their path lengths to the center point on the wall will be the same. We know that that means they will interfere constructively with one another. If we choose two points that are further in, but still the same distance from the middle of the slit, they will also have equal path lengths to the center point on the wall.

They will also interfere constructively with one another. So, we can see that there is a lot of constructive interference going on at that center point, in fact, there will be a major bright spot there because of it.

If we want to find a spot on the wall that is dark, we have to find where there is the most destructive interference. Figure of waves passing though single slit toward two different targets on opposite wall Because all of these pairs are the same distance apart across the slit, if we measure the path length from each pair to the same spot on the wall, each pair will have the same difference in path length. Figure of single-slit diffraction pattern If we compare single-slit diffraction to the double-slit interference pattern, the spots are much larger and more spread out.