In a Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light In a Young’s double-slit experiment, using mono chromatic light of wave length λ, the intensity of light at a point on the screen is I0, where the path difference between the interfering waves is λ. The path difference between the interfering waves at a point where the intensity is \(\frac{I_o}{2}\) , will be:at a point on the screen is I0, where the path difference between the interfering waves is λ. The path difference at a point where the intensity is \(2I_0\) will be:
The intensity \( I \) in Young's double-slit experiment is expressed as \( I = I_0 \cos^2(\frac{\phi}{2}) \), where \( \phi \) represents the phase difference between the interfering waves. The relationship between phase difference \( \phi \) and path difference \( \Delta x \) is given by \( \phi = \frac{2\pi}{\lambda} \Delta x \). It is established that when \( \phi = 2\pi \) (corresponding to \( \Delta x = \lambda \)), the intensity \( I = I_0 \). The objective is to determine \( \Delta x \) when \( I = \frac{I_0}{2} \).
Calculation proceeds as follows:
\( \frac{I_0}{2} = I_0 \cos^2(\frac{\phi}{2}) \Rightarrow \cos^2(\frac{\phi}{2}) = \frac{1}{2} \)
This equation yields \( \cos(\frac{\phi}{2}) = \pm\frac{1}{\sqrt{2}} \), leading to \( \frac{\phi}{2} = \frac{\pi}{4} \) or \( \frac{\phi}{2} = \frac{3\pi}{4} \).
Consequently, \( \phi = \frac{\pi}{2} \) or \( \phi = \frac{3\pi}{2} \).
Using the relation \( \phi = \frac{2\pi}{\lambda} \Delta x \), we find the path differences to be \( \Delta x = \frac{\lambda}{4} \) or \( \Delta x = \frac{3\lambda}{4} \).
The smallest non-zero path difference is \( \frac{\lambda}{4} \).
Therefore, the path difference corresponding to an intensity of \( \frac{I_o}{2} \) is \( \frac{\lambda}{4} \).