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**Hint:**When reflection occurs then there is phase change of $\pi $ radians between the incident ray and the reflected rays. The path difference between incident and the reflected ray is due to the phase change.

**Complete step by step solution:**

If $\theta $ is the phase difference between two waves (incident light wave and the reflected light ray) then the path difference between the two is calculated using formula,

$\Delta x=\left( \dfrac{\theta \lambda }{2\pi } \right)$

During reflection, $\theta =\pi $ radians

Hence, the path difference between the incident light and the reflected light can be calculated as,

$\Delta {{x}_{1}}=\left( \dfrac{\pi \lambda }{2\pi } \right)=\dfrac{\lambda }{2}$

At any point on the screen which is at distance of $y$ from the central point of the screen the path difference is given as,

$\Delta {{x}_{2}}=\dfrac{yd}{D}$

Where,

$\begin{align}

& d=\text{ slit width} \\

& D=\text{ distance of screen from the source} \\

\end{align}$

Then,

For constructive interference at point on the screen,

Phase difference $=N\lambda $

Where, $N=0,1,2\ldots \ldots $

Total path difference $=$$\Delta x=\Delta {{x}_{1}}+\Delta {{x}_{2}}$

$\begin{align}

& \Delta x=N\lambda \\

& \dfrac{yd}{D}+\dfrac{\lambda }{2}=N\lambda \\

& \dfrac{yd}{D}=N\lambda -\dfrac{\lambda }{2} \\

& =\left( 2N-1 \right)\dfrac{\lambda }{2}

\end{align}$

Therefore,

For the constructive interference at any point on the screen in Llyod’s single mirror experiment the path difference is equal to $\left( 2N-1 \right)\dfrac{\lambda }{2}$.

**Note:**1. Phase difference during reflection is $\pi $ radian

2. Phase difference during refraction depends on the path in which light is travelling. When light is moving from rarer medium to denser medium then the phase change is $\pi $

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