(S1P)2=D2+x2 and (S2P)2=(D−2λ)2+x2 So, (S1P)2−(S2P)2=D2+x2−[D2+4λ2−4Dλ+x2] =4Dλ−4λ2 (S1P+S2P)(S1P−S2P)=4Dλ−4λ2(D>>λ) Since, [D>>λ] ∴S1P=S2P and neglecting higher power of λ ∴2S1P(S1P−S2P)=4Dλ 2(S1P)(λ)=4Dλ or S1P=2D or √D2+x2=2D or x2=3D2 or x=√3D