4.4.10 The Young double-slit experiment Flashcards Preview

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Flashcards in 4.4.10 The Young double-slit experiment Deck (17):

Produce an equation connecting a, x and d. Define these values.

x = λD/a

where a<


What two conditions must be satisfied, in order to make measurements of the wavelength of light?

1. Light must be monochromatic (as this means that all of the light waves have the same wavelength);
2. There must be an accurate method f measuring small path differences (the small angle approximation);


What is the single most important thing about light from the double slit experiment?

That it is coherent; because all of it comes from the same, spherical source.


When does a dark fringe appear on the screen at the back of the double slit experiment?

When the path difference between the two different rays of light going there is half a wavelength out. This means the waves are in anti-phase, and so deconstructive superposition occurs. The waves are phi out of phase, 180 degrees.


Name three types of superposition and link these to the double slit experiment.

1. Constructive superposition - this occurs when the wavelengths of two different rays are whole wavelengths apart (in phase), appearing as bright fringes. This can be worked out from the path differences.
2. Total destructive superposition - this occurs when the wavelengths are in ant-phase and appear as dark fringes. Therefore, the waves have a path difference that is half of lambda plus some integer.
3. Partial destructive superposition - often confused with total destructive superposition. The waves are not in phase or antiphase - and so the double slit experiment fades into dark and light fringes, instead of being block (like a digital graph).


Define corpuscle.

People thought light was made up of these before Young's experiments in 1801 (including Newton). This was part of his theory of light.


Describe the 5 stage process that light goes through to get through the doubbe slit experiment.

1. Monochromatic light waves leave the same spherical light source X, and so are coherent.
2. Light waves reach the first slit which is comparable t the wavelength of the light, and so it diffracts.
3. Circular wave fronts reach the double slits. The waves are still coherent, and so diffract here.
4. Diffracted light passes through the double slits, overlaps and interferes - making dark and light fringes on that wall at the back.
5. The waves are progressive, so dark fringes can be seen where the waves arrive in antiphase and fringes where it is in phase.


What is special about the amplitude of the waves and the intensity of the light produces when they overlap?

The intensity of light is directly proportional to the amplitude squared, so in light fringes, the intensity of light will seem doubly as bright.


Name inaccuracies seen in diagrams of the double slit experiment.

1. The size of the double slits would be smaller, in order to compare to the wavelength of the light.
2. The wavelength is smaller, and o the light and dark fringes shown in diagrams would be closer in real life.


How should one attempt to determine the wavelength of light using the double-slit apparatus?

1. Done in a darkened room as light intensity is low (so hard to see fringes);
2. Use a filter so light is monochromatic, and fringes are sharper;
3. Measure across the fringes and divide this by the number of fringes in order to reduce % error.


Why is a balance needed between the slit--to-screen distance?

Because increasing it increases fringe separation, but reduces light intensity at the screen.


When does constructive interference occur?

When the path difference is a whole number of wavelengths.


Why is there a path difference?

Because two light rays which pass through the double slits travel different distances to get to the same point of the screen.


What is special about the maxima of a double slit experiment?

The maxima is the highest bright fringe on the screen that can be seen. The ray that goes through the bottom slit must travel one more wavelength than the wave that goes through the top slit. So the path difference must be one wavelength.


What gives the equation connecting lambda, a, D and x?

x = λD/a

where a<


What is special about interference?

All waves experience it (transverse or longitudinal) (coherent or non coherent);


What are the other applications of the double slit experiment?

It can also be applied to microwaves; where the wavelength is larger (around 3m). An amplifier can be connected to the transmitter, so the waves can be 'heard.'

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