Phase difference, $\phi$ = ${{2\pi } \over \lambda }$ $\times$ Path difference

When path difference is $\lambda$, then

$\phi$ = ${{2\pi } \over \lambda } \times \lambda$ = 2$\pi$

$\therefore$ I = 4I₀${\cos ^2}\left( {{{2\pi } \over 2}} \right)$ = 4I₀cos²($\pi$) = 4I₀ = K ....(1)

When path difference is , ${\lambda \over 4}$ then

$\phi$ = ${{2\pi } \over \lambda } \times {\lambda \over 4}$ = ${{\pi \over 2}}$

$\therefore$ I = 4I₀${\cos ^2}\left( {{{\pi } \over 4}} \right)$ = 2I₀ = ${K \over 2}$