Question:
Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength \(\lambda\) such that the first minima occurs directly in front of the slit (S1) ?
Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength \(\lambda\) such that the first minima occurs directly in front of the slit (S1) ?
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The point P is directly in front of S1, so S1P = 2d and S2P = sqrt((2d)^2 + d^2). Set the path difference equal to lambda/2 for the first minimum.
Updated On: Nov 5, 2026
- \(\frac{\lambda}{2(5-\sqrt2)}\)
- \(\frac{\lambda}{2(\sqrt5-2)}\)
- \(\frac{\lambda}{(5-\sqrt2)}\)
- \(\frac{\lambda}{(\sqrt5-2)}\)
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The Correct Option is B
Solution and Explanation
From the figure the slit separation is \(d\) and the screen is at distance \(2d\). P is directly in front of \(S_1\).
\(S_1P=2d\).
\(S_2P=\sqrt{(2d)^2+d^2}=\sqrt5\,d\).
First minimum: path difference \(=\frac{\lambda}{2}\).
\((\sqrt5-2)d=\frac{\lambda}{2}\), so \(d=\frac{\lambda}{2(\sqrt5-2)}\).
Answer: (2) \(\frac{\lambda}{2(\sqrt5-2)}\)
\(S_1P=2d\).
\(S_2P=\sqrt{(2d)^2+d^2}=\sqrt5\,d\).
First minimum: path difference \(=\frac{\lambda}{2}\).
\((\sqrt5-2)d=\frac{\lambda}{2}\), so \(d=\frac{\lambda}{2(\sqrt5-2)}\).
Answer: (2) \(\frac{\lambda}{2(\sqrt5-2)}\)
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