Question

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:

• A. \(\frac{\lambda}{2}\)
• B. \(\frac{\lambda}{4}\)
• C. λ
• D. 2λ

Answer 1

The Correct Option is B

In Young’s double-slit experiment, the intensity \( I \) at a point on the screen is given by the formula \( I = I_0 \cos^2(\frac{\phi}{2}) \),

where \( \phi \) is the phase difference between the interfering waves.
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by \( \phi = \frac{2\pi}{\lambda} \Delta x \).
Given that at \( \phi = 2\pi \), \( I = I_0 \) when \( \Delta x = \lambda \).
Now, we need to find \( \Delta x \) when \( I = \frac{I_0}{2} \).

Let's calculate:

\( \frac{I_0}{2} = I_0 \cos^2(\frac{\phi}{2}) \Rightarrow \cos^2(\frac{\phi}{2}) = \frac{1}{2} \)

This implies \( \cos(\frac{\phi}{2}) = \pm\frac{1}{\sqrt{2}} \), giving \( \frac{\phi}{2} = \frac{\pi}{4} \) or \( \frac{3\pi}{4} \).

Hence \( \phi = \frac{\pi}{2} \) or \( \phi = \frac{3\pi}{2} \).

From \( \phi = \frac{2\pi}{\lambda} \Delta x \), we find \( \Delta x = \frac{\lambda}{4} \) or \( \Delta x = \frac{3\lambda}{4} \).

The smallest non-zero path difference is \( \frac{\lambda}{4} \).

Therefore, the path difference at a point where the intensity is \(\frac{I_o}{2}\) is \( \frac{\lambda}{4} \).