In figure $A$ and $B$ represent the edges of the slit $AB$ of width $a$ and $C$ represents the midpoint of the slit. For the first minimum at $P$,
$a \sin \theta=\lambda$ ...... $(i)$
where $\lambda$ is the wavelength of light. The path difference between the wavelets from $A$ to $C$ is
$\Delta x=\frac{a}{2} \sin \theta=\frac{1}{2}(a \sin \theta)=\frac{\lambda}{2}$ (using $(i)$)
The corresponding phase difference $\Delta \phi$ is
$\Delta \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}=\pi$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
