An Explanation of the Intensity of Young’s Double Slit Experiment

In 1801, Thomas Young conducted an experiment that demonstrated the interference pattern of light waves. This famous experiment is known as the Young’s Double Slit Experiment. In this experiment, monochromatic light is passed through two slits, and the intensity of the light is measured on a screen behind the slits. What Young found was that when the phase difference between two waves was zero (meaning they were in sync), the intensity was at its maximum. When the path difference between two waves was at its maximum, the intensity was at its minimum. This phenomenon can be explained by interference patterns.

What is the intensity of Young’s Double Slit Experiment?

The intensity of light is a measure of the amount of power that is transmitted through a given area. In the case of Young’s Double Slit Experiment, the intensity is affected by the phase difference and path difference between the two waves.

What is the phase difference?

The phase difference is the difference in the position of the peaks and troughs of the two waves. In the case of Young’s Double Slit Experiment, the phase difference is created by the different path lengths taken by the two waves.

What is the path difference?

The path difference is the difference in the distance traveled by the two waves. In the case of Young’s Double Slit Experiment, the path difference is created by the different refractive indices of the two materials.

What is the maximum intensity of Young’s Double Slit Experiment?

The intensity of light is defined as the power per unit area. The maximum intensity occurs when the phase difference between the two waves is zero. When the path difference is equal to one wavelength, the intensity is at its maximum. The intensity decreases as the path difference increases. When the path difference is equal to half a wavelength, the intensity is at its minimum.

How can the intensity of light be increased?

The intensity of light can be increased by using a higher power source. However, the intensity will eventually reach a saturation point where adding more power will not increase the intensity. The saturation point is determined by the phase difference between the two waves. When the phase difference is zero, the intensity is at its maximum. When the phase difference is equal to half a wavelength, the intensity is at its minimum.

The intensity of light can also be increased by decreasing the path difference. The intensity will reach a maximum when the path difference is equal to one wavelength. As the path difference decreases, the intensity will increase. When the path difference is equal to half a wavelength, the intensity will be at its minimum.

The intensity of light can also be increased by increasing the number of slits. The more the slits, the more light that can pass through. However, at a certain point, the intensity will reach a saturation point where adding more slits will not increase the intensity.

What is the maximum intensity formula?

The formula for the intensity of light at a certain point is:

intensity = I0 * (sin( phase difference / (path difference)))^-2

Where:

I0 is the intensity of the light source

The phase difference is the difference in the phase of the waves reaching different parts of the screen

The path difference is the difference in path length between the waves reaching different parts of the screen

Two is a constant

So, intensity is directly proportional to I0 and inversely proportional to the square of the path difference. The intensity is also directly proportional to the sine of the phase difference.

Now let’s look at an example. In Young’s double-slit experiment, light from a monochromatic source (i.e. one wavelength) is shone through two slits. The light waves passing through the two slits will have a phase difference and so will produce interference fringes on a screen.

The intensity of the fringes will be highest where the path difference is an integer multiple of the wavelength (i.e. when the waves are in phase) and lowest where the path difference is an odd multiple of a half-wavelength (i.e. when the waves are out of phase).

intensity = I0 * (sin( phase difference / (wavelength)))^-2

intensity =  I0* (sin(0))^-2

intensity = I0 * (0)^-2

intensity = I0 * (1)

intensity = I0

This formula tells us that the intensity is directly proportional to I0, and so the intensity of the fringes will be directly proportional to the intensity of the light source.

Now let’s look at what happens when the path difference is an odd multiple of a half wavelength.

intensity =  I0* (sin( phase difference / (wavelength)))^-2

intensity = I0 * (sin((pi)/wavelength))^-2

intensity = I0 * (sin(pi))^-2

intensity = I0 * (-one)^-2

intensity = I0 * (1)

intensity = I0

Again, the intensity is directly proportional to I0. So the intensity of the fringes will be the same, regardless of whether the path difference is an odd multiple of a half-wavelength or not.

Conclusion

When the intensity is plotted as a function of the phase difference, a clear sinusoidal wave is seen. This intensity variation is due to the path difference between the two slits; when the path difference is an integer multiple of the wavelength (i.e. when the phase difference is 0 or pi), constructive interference occurs and the intensity is at a maximum. When the path difference is an odd multiple of a quarter wavelength (i.e. when the phase difference is pi/ two or -pi/two), destructive interference occurs and the intensity is at a minimum. In all other cases, the intensity will be somewhere in between these two extremes.