Interference Intensity Difference Calculation

This solution details the calculation for the difference between maximum and minimum intensities in an interference pattern formed by superimposing two coherent light beams.

Intensity-Amplitude Relationship

The intensity ($I$) of a light wave is directly proportional to the square of its amplitude ($A$). The relationship can be expressed as:

$I = k A^2$

where $k$ is the constant of proportionality. From this, we know that the amplitude is proportional to the square root of the intensity: $A \propto \sqrt{I}$.

Deriving Amplitudes

We are given two coherent monochromatic light beams with intensities $I_1 = 4I$ and $I_2 = 9I$. Let their respective amplitudes be $A_1$ and $A_2$.

• Using the relationship $A \propto \sqrt{I}$: $A_1 \propto \sqrt{4I} = 2\sqrt{I}$
• $A_2 \propto \sqrt{9I} = 3\sqrt{I}$

Let $A_{base} = \sqrt{I}$. We can represent the amplitudes as $A_1 = 2A_{base}$ and $A_2 = 3A_{base}$ for simplicity, understanding that these are proportional values.

Calculating Resultant Amplitudes

In an interference pattern, the maximum resultant amplitude ($A_{max}$) occurs during constructive interference, and the minimum resultant amplitude ($A_{min}$) occurs during destructive interference.

• Maximum Amplitude: $A_{max} = A_1 + A_2 = 2A_{base} + 3A_{base} = 5A_{base}$
• Minimum Amplitude: $A_{min} = |A_2 - A_1| = |3A_{base} - 2A_{base}| = |A_{base}| = A_{base}$

Calculating Resultant Intensities

The resulting maximum intensity ($I_{max}$) and minimum intensity ($I_{min}$) are proportional to the squares of the maximum and minimum amplitudes, respectively.

• $I_{max} \propto A_{max}^2 = (5A_{base})^2 = 25 A_{base}^2$
• $I_{min} \propto A_{min}^2 = (A_{base})^2 = 1 A_{base}^2$

Since $A_{base}^2$ is proportional to $I$ (from $A \propto \sqrt{I}$), we can express the maximum and minimum intensities in terms of $I$: $I_{max} = 25I$ $I_{min} = 1I = I$

Finding the Intensity Difference

The difference between the maximum and minimum intensities is calculated as:

$ \Delta I = I_{max} - I_{min} = 25I - I = 24I $

Determining the Value of x

The question states that the difference between the maximum and minimum intensities is $xI$. By comparing our calculated difference with the given expression:

$ xI = 24I $

Dividing both sides by $I$ (assuming $I \neq 0$), we find the value of $x$:

$ x = 24 $