Question

Find the intensity at a point on the screen in Young’s double slit experiment, at which the interfering waves of intensity \( I_0 \) each, have a path difference of (i) \( \frac{\lambda}{3} \), and (ii) \( \frac{\lambda}{2} \).

Hint:

Use \( I = 2I_0(1 + \cos\phi) \) when two coherent sources of equal intensity interfere. Convert path difference to phase angle using \( \phi = \frac{2\pi}{\lambda} \cdot \Delta x \).

Answer 1

The intensity at a point in Young’s double slit experiment is given by: \[ I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi \] If \( I_1 = I_2 = I_0 \), and \( \phi = \frac{2\pi}{\lambda} \cdot \text{path difference} \), then: \[ I = 2I_0(1 + \cos\phi) \] (i) For \( \text{path difference} = \frac{\lambda}{3} \): \[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \Rightarrow I = 2I_0\left(1 + \cos\frac{2\pi}{3} \right) = 2I_0\left(1 - \frac{1}{2} \right) = I_0 \] (ii) For \( \text{path difference} = \frac{\lambda}{2} \): \[ \phi = \pi \Rightarrow I = 2I_0(1 + \cos\pi) = 2I_0(1 - 1) = 0 \]