In Young’s double-slit experiment, let \( \beta \) be the fringe width and \( I_0 \) the intensity at the central bright fringe. At a distance x from the central bright fringe, the intensity will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( I_0\cos^{2}(\pi x/\beta) \)
For two coherent sources of equal intensity \( I_1 \), the intensity at a point of phase difference \( \phi \) is \( I=4I_1\cos^2(\phi/2) \). At the central bright fringe \( \phi=0 \), so \( I_0=4I_1 \) and \( I=I_0\cos^2(\phi/2) \).
At a distance \( x \) from the central fringe, the path difference is \( xd/D \). With the fringe width \( \beta=\lambda D/d \), the phase difference is:
\( \begin{aligned} \phi &= \frac{2\pi}{\lambda}\cdot\frac{xd}{D} \\ &= \frac{2\pi x}{\beta} \end{aligned} \)
The intensity at \( x \) is then:
\( \begin{aligned} I &= I_0\cos^2\left(\frac{\phi}{2}\right) \\ &= I_0\cos^2\left(\frac{\pi x}{\beta}\right) \end{aligned} \)
At \( x=\beta \) this gives \( I_0 \) again, at the next bright fringe.