Question

In Young's double slit experiment, the intensity of each source is $I_0$. The resultant intensity at a point on screen, where path difference is $\frac{3\lambda }{4}$, is,

• A. $0$
• B. $I_0$
• C. $2I_0$
• D. $3I_0$

Answer 1

The correct option is C $2I_0$

The phase difference is given by,
$\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}=\frac{3\pi }{2}$

The resultant intensity of interference of light waves from two coherent sources, having phase difference $\varphi$, is given by,
$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

As, $\varphi =\frac{3\pi }{2}$ and $\cos \varphi =0$

$\therefore I_R=I_0+I_0=2I_0$

Hence, $\left(C\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{For better understanding of relation between phase difference and path difference during interference of light waves.}\end{array}$