$\triangle\text{S}_1\text{PO}$ and $\triangle\text{S}_2\text{PO}$ are right-angled triangle
So, $(\text{S}_1\text{P})^2-(\text{S}_2\text{P})^2$$=\big[\text{D}^2+\text{x}^2\big]-\Big[(\text{D}-2\lambda)^2+\text{x}^2\Big]^2$
$=4\lambda\text{D}+4\lambda^2=4\lambda\text{D}$
$\Big(\lambda^2$ is small and can be neglected$\Big)$
$\Rightarrow(\text{S}_1\text{P}+\text{S}_2\text{P})(\text{S}_1\text{P}-\text{S}_2\text{P})=4\lambda\text{D}$
$\Rightarrow(\text{S}_1\text{P}-\text{S}_2\text{P})=\frac{4\lambda\text{D}}{(\text{S}_1\text{P}+\text{S}_2\text{P})}$
$\Rightarrow\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}$
For constructive interference, path difference $=\text{n}^\lambda$ So, $\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}=\text{n}\lambda$$\Rightarrow\frac{2\text{D}}{\sqrt{\text{x}^2+\text{D}^2}}=\text{n}$
$\Rightarrow\text{n}^2\big(\text{x}^2+\text{D}^2\big)=4\text{D}^2$
$\Rightarrow\text{n}^2\text{x}^2+\text{n}^2\text{D}^2=4\text{D}^2$
$\Rightarrow\text{n}^2\text{x}^2=4\text{D}^2-\text{n}^2\text{D}^2$
$\Rightarrow\text{n}^2\text{x}^2=\text{D}^2(4-\text{n}^2)$
$\Rightarrow\text{x}=\frac{\text{D}}{\text{n}}\sqrt{4-\text{n}^2}$
When n = 1,$\text{x}=\sqrt{3}\text{D}$ (1st order).
When n = 2,$\text{x}=0$ (2nd order).
So when $\text{x}=\sqrt{3}\text{D},$ the intensity at P is equal to the intensity at O.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
