Minimum and maximum intensity in Young's double slit experiment

It is very difficult to find the runway for aircraft in bad weather.

The aircraft can not even land properly in bad weather.

The interference from two coherent sources is used to make guiding beacons for aircraft to land properly.

The concept of interference from two coherent sources has been explained by Young.

We must have an idea of how these waves superimpose after passing through two different coherent sources and also the resultant intensity of these waves.

Let's discuss the intensity of the superimposing waves.

The intensity of a wave is the power transferred per unit area, where the area is measured on a plane perpendicular to the direction of propagation of energy.

The intensity of a wave can be determined if the amplitude, $A$ of the wave is known.

The intensity of a wave is directly proportional to the square of the amplitude.

Similarly, the intensity of the resultant wave, $I_R$ is directly proportional to the resultant amplitude, $R$.

Consider two waves of amplitude $A_1$ and $A_2$ having a phase difference of $\varphi$.

The resultant amplitude, $R$ of both waves will be given by,

Solving the resultant amplitude equation for intensity,

After rearranging the terms, the generic equation for intensity becomes,

The resultant intensity will change by changing the values of the phase angle, $\varphi$ otherwise it will remain the same.

If phase angle, $\varphi$ is zero then the equation becomes,

Now, suppose waves of equal intensities are superimposing with zero phase angle, then, the resultant intensity becomes, maximum.

Now if the phase difference between the waves is $\pi$,

Again consider waves having the same amplitudes with $\pi$ phase difference, then, resultant intensity becomes zero.

Now let's discuss the minimum and maximum intensity of YDSE

Consider a wave having phase angle of $2\pi$ and wavelength, $\lambda$.

Let's assume a particle $P$ with $\varphi$ phase difference on the wave at a length $x$ from the origin.

The phase difference, $\varphi$ can be evaluated as,

Let's calculate maximum resultant amplitude, $R$ of the waves at $\varphi =0$.

Now we will find the minimum resultant amplitude, $R$ of the wave at $\varphi =\pi$.

Since we know the maximum and minimum amplitude, we can find out the ratio of maximum and minimum intensity.

The intensity is always proportional to the square of the amplitude. So the ratio of maximum and minimum intensity becomes,

After replacing the value of the minimum and maximum amplitude, the ratio of intensity becomes,

Revision

The expression for the resultant intensity of two waves can be derived as,

The expression for resultant intensity at $0$ phase difference becomes,

The expression for resultant intensity at $\pi$ phase difference becomes,

The expression for the ratio of minimum and maximum intensity of waves can be derived as,

The End