Question

The phase difference between the two superimposing waves that give rise to a bright spot in a Young’s double-slit experiment is (\(n\) is an integer) :

• A. \(2n\pi\)
• B. \(2n\pi + \frac{\pi}{4}\)
• C. \(2n\pi + \frac{\pi}{2}\)
• D. \(2n\pi + \pi\)

Hint:

Remember the simple rules for interference: - Bright spot (Constructive): Phase difference \(\phi = 2n\pi\); Path difference \(\Delta x = n\lambda\). - Dark spot (Destructive): Phase difference \(\phi = (2n+1)\pi\); Path difference \(\Delta x = (2n+1)\frac{\lambda}{2}\).

Answer 1

The Correct Option is A

Concept: Interference is a phenomenon where two coherent light waves superimpose to produce a resultant wave of greater or lower intensity.
• Constructive Interference: Occurs when the waves meet in phase, leading to maximum intensity (a bright spot).
• Destructive Interference: Occurs when the waves meet completely out of phase, leading to minimum intensity (a dark spot). The resultant intensity \(I\) of two interfering waves with intensities \(I_1\) and \(I_2\) and a phase difference \(\phi\) is expressed as: \[ I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi \]

Step 1: Mathematical condition for maximum intensity (bright fringe).
To get a bright spot, the resultant intensity \(I\) must be maximized. This directly requires the interference term containing \(\cos\phi\) to take its maximum possible mathematical value: \[ \cos\phi = +1 \] We know from trigonometry that the cosine function achieves a value of \(+1\) at all integer multiples of \(2\pi\): \[ \phi = 0, \pm 2\pi, \pm 4\pi, \pm 6\pi, \ldots \] This sequence can be expressed in general algebraic form as: \[ \phi = 2n\pi \] where \(n \in \mathbb{Z}\) (\(n = 0, 1, 2, 3, \ldots\)).

Step 2: Corresponding path difference link.
The relationship between the phase difference (\(\phi\)) and path difference (\(\Delta x\)) is given by: \[ \phi = \frac{2\pi}{\lambda} \Delta x \] Substituting the bright spot condition \(\phi = 2n\pi\): \[ 2n\pi = \frac{2\pi}{\lambda} \Delta x \quad \Rightarrow \quad \Delta x = n\lambda \] Thus, constructive interference happens when the path difference is an integral multiple of the wavelength, which requires the phase difference to be an even integral multiple of \(\pi\), i.e., \(\phi = 2n\pi\). This aligns perfectly with Option (A).