In Young's double slit experiment the intensity at a point on the screen is $K$, where path difference is…
In Young's double slit experiment the intensity at a point on the screen is $K$, where path difference is $\lambda$. What will be the intensity at the point where path difference is $\frac{\lambda}{4}$ ?
$\frac{K}{4}$
$\frac{K}{2}$
$K$
0
Solution
$I=2 I_0(1+\cos \phi)$
$\phi=\frac{2 \pi x}{\lambda}$, where $x$ is the path as difference.
$\therefore K=2 I_0\left(1+\cos \left(\frac{2 \pi}{\lambda} \times \lambda\right)\right)=4 I_0$
Now, for path difference $\frac{\lambda}{4}$
$\begin{aligned}
& K^{\prime}=2 I_0\left(1+\cos \left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\right)\right)=2 I_0 \\
& \therefore K^{\prime}=\frac{K}{2}
\end{aligned}$