Question

In a Young's double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{2} \) on the screen. Find the intensity at this point.

Hint:

When the path difference between two waves is \( \frac{\lambda}{2} \), the waves are in destructive interference, and the intensity at that point will be zero.

Answer 1

In a Young's double-slit experiment, the intensity at a point where the two waves interfere is given by the equation: \[ I = I_0 \left( 1 + \cos \delta \right) \] where: - \( I_0 \) is the intensity of each individual wave, - \( \delta \) is the phase difference between the two waves. The phase difference \( \delta \) is related to the path difference \( \Delta \) by the equation: \[ \delta = \frac{2\pi \Delta}{\lambda} \] Given that the path difference is \( \Delta = \frac{\lambda}{2} \), we can substitute into the equation for \( \delta \): \[ \delta = \frac{2\pi \times \frac{\lambda}{2}}{\lambda} = \pi \] Now, substitute \( \delta = \pi \) into the intensity formula: \[ I = I_0 \left( 1 + \cos \pi \right) = I_0 \left( 1 - 1 \right) = 0 \] Thus, the intensity at this point is zero.