Question

The condition for constructive interference in Lloyd’s single mirror experiment is the path difference which is equal to
A. $N\lambda $
B. $(2N-1)\dfrac{\lambda }{2}$
C. $(N-1)\dfrac{\lambda }{2}$
D. $\dfrac{\lambda }{2(2N-1)}$

Answer 1

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 $