Ta có: \(\lambda=\dfrac{v}{f}=\dfrac{2}{10}=0,2m\)
mà cùng pha nên \(-\dfrac{S_1S_2}{\lambda}\le k\le\dfrac{S_1S_2}{\lambda}\)\(\Rightarrow-2\le k\le2\)
\(d_2-d_1=k\lambda\Leftrightarrow d_1=d_2-k\lambda\), để \(d_1\) lớn nhất suy ra k=1 (k\(\ne\)0)
\(\Rightarrow d_2-d_1=1.\lambda\Rightarrow\sqrt{0,4^2+d_1^2}-d_1=0,2\)
\(\Rightarrow d_1=0,3m\)