Physics - Wave Optics Question with Solution | TestHub

Initially, path difference at$\mathrm{P}$between two waves reaching from$\mathrm{S}$and$\mathrm{S}\text{'}$is$2\mathrm{h}$



Therefore, for maximum intensity at$\mathrm{P}$:

$2\text{h}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{\lambda }\text{ ....... (i) }$

Now, let the sources$\mathrm{S}$is moved by a distance$\mathrm{x}$, the path difference will be
$2\text{h}+2x \text{ or }2\text{h}-2x$So, for displaced by x ( away or towards the mirror) then maximum intensity at$\mathrm{P}$,

$2\text{h}+2\mathrm{x}=\left(\mathrm{n}+1\mathrm{-}\frac{1}{2}\right)\mathrm{\lambda } \; \; \; \; \; \; \; \text{....... (ii)}$

$2\text{h}-2\mathrm{x}=\left(\mathrm{n}-1-\frac{1}{2}\right)\mathrm{\lambda }\text{ .......(iii) }$


Solving Eqs. (i) and (ii) or Eqs. (i) and (iii), we get,

$2\text{x}=\left(\text{n+}\frac{\text{1}}{\text{2}}\right)\mathrm{\lambda }-\left(\text{n}-\frac{\mathrm{1}}{\text{2}}\right)\mathrm{\lambda }$
$=\mathrm{\lambda }$
$\text{x}=\frac{\mathrm{\lambda }}{2}$
$\text{x}=\frac{\mathrm{\lambda }}{2}=\frac{\text{600}}{\text{2}}=\text{300 nm}$
NoteHere, we have taken the condition of maximum intensity at P as:

Path difference$\Delta \mathrm{x}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{\lambda }$

Because the reflected beam suffers a phase difference of$\pi$.