In Young's double-slit experiment (YDSE), when two coherent waves of equal intensity $I_o$ from each slit interfere, the resultant intensity $I$ at any point on the screen depends on the phase difference ($\delta$) between the waves arriving at that point.
The formula for the resultant intensity is:
$I = I_o + I_o + 2\sqrt{I_o \cdot I_o} \cos(\delta)$
This formula simplifies to:
$I = 2I_o (1 + \cos(\delta))$
To find the intensity value of $2I_o$, we set $I = 2I_o$ in the intensity formula:
$2I_o = 2I_o (1 + \cos(\delta))$
Dividing both sides by $2I_o$ yields:
$1 = 1 + \cos(\delta)$
This equation implies that:
$\cos(\delta) = 0$
The condition $\cos(\delta) = 0$ occurs when the phase difference $\delta$ is $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ (or any odd multiple thereof). This corresponds to specific points on the screen where the path difference is $\Delta r = \frac{\lambda}{4}$ or $\Delta r = \frac{3\lambda}{4}$. At these points, the intensity is $2I_o$. While the phrasing "in front of one of the slits" can be ambiguous, the intensity value $2I_o$ is achieved under the condition $\cos(\delta) = 0$.
Therefore, the intensity on the screen is $2I_o$.
As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.
If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.