91影视

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

Assume that the slit width is much smaller than the wavelength of the light, therefore the diffraction envelope can be ignored.

The slits are equidistantly spaced means that the phase angle between the coming light and each slit is$蠒$. Hence, the electric field at each slit can be written as follows:

$$\begin{gathered} E_1=E_0\sin \left(蝇t\right) \\ {E}_2=E_0\sin \left(蝇t+蠒\right) \\ {E}_3=E_0\sin \left(蝇t+2蠒\right) \end{gathered}$$

Which can be represented in the phasor image given below.

From the phasor diagram, the net electric field at any point on the screen is given by:

$$\begin{gathered} E=E_1+E_2+E_3 \\ =E_0+E_0\left(\cos 蠒\right)+E_0\left(\cos 蠒\right) \\ =E_0\left(1+2\cos 蠒\right) \end{gathered}$$

It is known that the intensity is proportional to the square of the electric field. So,

$I鈭�E^2鈬�I=A{E}^2$

Substitute in the above equation to get $I=A{\left(E_0\left(1+2\cos 蠒\right)\right)}^2$. Now, the intensity is maximum when the phase difference is zero between the electric field. So,

$$\begin{gathered} I=A{\left(E_0\left(1+2\cos 蠒\right)\right)}^2 \\ {I}_m=A{\left(E_0\left(1+2\cos 0\right)\right)}^2 \\ {I}_m=A{\left(E_0\left(3\right)\right)}^2 \\ {I}_m=9A{E}_0^2 \end{gathered}$$

The value of$A$ from the above equation, we get$A=\frac{I_m}{9{E_0}^2}$ . Therefore, the intensity of the electric field will become:

$$\begin{gathered} I=\frac{I_m}{9{E_0}^2}{\left(E_0\left(1+2\cos 蠒\right)\right)}^2 \\ I=\frac{I_m}{9}\left(1+4\cos 蠒+4{\cos }^2蠒\right) \end{gathered}$$

Hence, it is proved that the intensity is $I=\frac{1}{9}{I}_m\left(1+4\cos 蠒+4{\cos }^2蠒\right)$.