Question

If you double the width of a single slit, the intensity of the central maximum of the diffraction pattern increases by a factor of 4, even though the energy passing through the slit only doubles. Explain this quantitatively

Short answer

It is proved that the intensity of the intensities is$4:1$

Step-by-step solution

Identification of given data

The given data can be listed below,

• The width of the single slit is,$d\text{'}=2d$
• The intensity of the central maxima is,$I\text{'}\left(\theta \right)=4I\left(\theta \right)$

Here,$d\text{and}I\left(\theta \right)$ are the initial slit width and initial intensity of the wave.

Diffraction

When light beams pass through an opaque object or an aperture with a diameter that is less than or equal to the wavelength, the result is a physical phenomenon known as diffraction.

Explanation

The intensity of diffraction pattern at an angle is given by,

$I\left(\theta \right)=I_m{\left(\frac{\sin \alpha }{\alpha }\right)}^2$

Here,$I_m$ is the initial intensity of diffraction pattern.

The degree of the intensity is proportional to the square of the number of phasors, if the slid width is doubled, we divide the original slit into N slits (or, N phasors). At the central maximum, all phasors are in the same direction and have the same amplitude.

The diffraction pattern’s intensity is given as,

$$\begin{gathered} I\left(\theta \right)\propto N^2 \\ I\left(\theta \right)=K{N}^2 \end{gathered}$$

Here, $k$is the constant of proportionality,

Now if the slit width is double, the number of phasors N is given by,

$$\begin{gathered} I\text{'}\left(\theta \right)\propto {\left(2N\right)}^2 \\ I\text{'}\left(\theta \right)=4N^2 \end{gathered}$$

The ratio of intensity is given by,

$$\begin{gathered} \frac{I\text{'}\left(\theta \right)}{I\left(\theta \right)}=\frac{4I\left(\theta \right)}{I\left(\theta \right)} \\ =4:1 \end{gathered}$$

Thus, it is proved that the intensity of the intensities is $4:1$