Path Difference in Young Double Slit Experiment for Equal Intensity

To find the path difference between the interfering waves at point $P$, we use the expression for the resultant intensity of two coherent light waves of equal individual intensity $I_0$:

$I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right)$

where $\phi$ is the phase difference between the two waves.

It is given that the resultant intensity $I$ at point $P$ is equal to the intensity of an individual slit, i.e., $I = I_0$:

$I_0 = 4 I_0 \cos^2\left(\frac{\phi}{2}\right)$

Dividing both sides by $I_0$:

$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4}$

Taking the square root on both sides:

$\cos\left(\frac{\phi}{2}\right) = \pm \frac{1}{2}$

Thus, the phase difference $\phi$ can be written as:

$\frac{\phi}{2} = n\pi \pm \frac{\pi}{3} \quad (n \in \mathbb{Z})$ $\phi = 2n\pi \pm \frac{2\pi}{3}$

The relation between phase difference $\phi$ and path difference $\Delta x$ is given by:

$\Delta x = \frac{\lambda}{2\pi} \phi$

1. For the first possible value of phase difference ($n = 0$, taking the positive sign): $\phi = \frac{2\pi}{3}$ $\Delta x = \frac{\lambda}{2\pi} \left(\frac{2\pi}{3}\right) = \frac{\lambda}{3}$ Substituting $\lambda = 6000\text{ \AA}$: $\Delta x = \frac{6000\text{ \AA}}{3} = 2000\text{ \AA}$

2. For the next possible value of phase difference ($n = 1$, taking the negative sign): $\phi = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$ $\Delta x = \frac{\lambda}{2\pi} \left(\frac{4\pi}{3}\right) = \frac{2\lambda}{3}$ Substituting $\lambda = 6000\text{ \AA}$: $\Delta x = \frac{2 \times 6000\text{ \AA}}{3} = 4000\text{ \AA}$

Thus, both $2000\text{ \AA}$ and $4000\text{ \AA}$ represent valid path differences for the given condition.

Correct Options: A ($4000\text{ \AA}$) and C ($2000\text{ \AA}$)