A narrow slit S transmitting light of wavelength λ is positioned a...

In Lloyd's mirror experiment, the setup is equivalent to Young's Double Slit Experiment (YDSE). The real source $S$ and its virtual image $S'$ formed by the plane mirror act as two coherent sources.

The distance between the real source $S$ and the virtual source $S'$ is $a = 2d$.

The path difference between the direct ray and the reflected ray at a distance $y$ from $O$ on the screen is given by:
$\Delta x = \dfrac{y \cdot a}{D} = \dfrac{y(2d)}{D}$

When light reflects from the denser medium (the mirror), it undergoes a phase change of $\pi$, which is equivalent to an additional path difference of $\dfrac{\lambda}{2}$.

Therefore, the condition for constructive interference (maxima) is that the geometric path difference must be an odd multiple of $\dfrac{\lambda}{2}$:
$\Delta x = \left(n + \dfrac{1}{2}\right)\lambda$

For the first maximum, $n = 0$, so the geometric path difference is:
$\Delta x = \dfrac{\lambda}{2}$

Equating the two expressions for the path difference:
$\dfrac{y(2d)}{D} = \dfrac{\lambda}{2}$

Solving for $y$, we get the distance of the first maximum from $O$:
$y = \dfrac{D\lambda}{4d}$