Two sources of sound vibrate at the same wavelength \(\lambda\) and in phase. The setup is similar to the double slit experiment of light with one major variation, that is \(d = 2D\) (in YDSE, usually \(D \gg d\)). Will fringe width be $\frac{\lambda D}{d}$ or \(\frac{\lambda}{2}\) for \(y \ll D\)? Given that symbols have their usual meaning.
Comprehensive Solution & Analysis
At first glance, this question looks like a simple plug-and-chug problem. If you substitute \(d = 2D\) into the standard Young's Double-Slit Experiment (YDSE) fringe width formula \(w = \frac{\lambda D}{d}\), you get:
\(w = \frac{\lambda D}{2D} = \frac{\lambda}{2}\)
Algebraically, the options \(\lambda \cdot \frac{D}{d}\) and \(\frac{\lambda}{2}\) represent the exact same value. However, neither of them is correct for this setup. Here is why and how to solve it properly.
1. Why the Standard Formula Fails Here
The standard formula \(w = \frac{\lambda D}{d}\) relies strictly on the paraxial approximation, which assumes that the distance to the screen (\(D\)) is vastly larger than the source separation (\(d\)) (\(D \gg d\)). This assumption allows us to treat the rays traveling from both sources to the screen as nearly parallel, simplifying the path difference to \(\Delta x \approx \frac{yd}{D}\).
In this problem, we are explicitly given that \(d = 2D\). Because \(D\) is comparable to \(d\), the paraxial approximation completely breaks down, and we must return to the exact geometry of the wave paths.
2. Deriving the Exact Path Difference
Using the exact geometric path difference (\(\Delta x\)) and substituting \(d = 2D\) (so \(\frac{d}{2} = D\)), we step through the approximation for \(y \ll D\) (or equivalently, \(2D^2 \gg y^2\)):
\(\begin{aligned} \Delta x &= \sqrt{{D^2} + {{\left( {y + \frac{d}{2}} \right)}^2}} - \sqrt{{D^2} + {{\left( {y - \frac{d}{2}} \right)}^2}} \\ &= \sqrt{{D^2} + {{\left( {y + D} \right)}^2}} - \sqrt{{D^2} + {{\left( {y - D} \right)}^2}} \\ &= \sqrt{2{D^2} + {y^2} + 2yD} - \sqrt{2{D^2} + {y^2} - 2yD} \\ &\approx \sqrt{2{D^2} + 2yD} - \sqrt{2{D^2} - 2yD} \quad (2D^2 \gg y^2)\\ &= D\sqrt{2} \left\{ {{\left( {1 + \frac{y}{D}} \right)}^{1/2}} - {{\left( {1 - \frac{y}{D}} \right)}^{1/2}} \right\} \\ &\approx D\sqrt{2} \left\{ {\left( {1 + \frac{y}{2D}} \right) - \left( {1 - \frac{y}{2D}} \right)} \right\} \\ &= y\sqrt{2} \end{aligned}\)
3. Calculating the True Fringe Width
For constructive interference (maxima), the path difference must equal an integer multiple of the wavelength (\(\lambda\)):
\(y\sqrt{2} = n\lambda \implies y_n = \frac{n\lambda}{\sqrt{2}}\)
The fringe width (\(w\)) is defined as the spatial separation between two successive maxima (\(y_{n+1} - y_n\)):
\(w = \frac{(n+1)\lambda}{\sqrt{2}} - \frac{n\lambda}{\sqrt{2}} = \frac{\lambda}{\sqrt{2}}\)