Hereafter I need all ans in English

The phase difference ($\Delta \phi$) between two waves is related to their path difference ($\Delta x$) by the formula:

where $\lambda$ is the wavelength of light.

For a bright fringe (constructive interference), the path difference is an integer multiple of the wavelength:

where $m$ is the order of the bright fringe ($m = 0, 1, 2, \dots$).

The central bright fringe corresponds to $m=0$.
The first bright fringe corresponds to $m=1$.
The second bright fringe corresponds to $m=2$.
The third bright fringe corresponds to $m=3$.
So, for the third bright fringe, $m=3$.

Using $m=3$, the path difference is:

Substitute $\Delta x = 3\lambda$ into the phase difference formula:

The wavelength $\lambda = 6000 \text{ Å}$ is given but is not required for this calculation as it cancels out.

The final answer is $\text{(d) } 6\pi$