# NEET Physics Wave Optics Questions Solved

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(a) Derive an expression for path difference in Young's double slit experiment and obtain  the conditions for constructive and destructive at a point on the screen.

(b) The intensity at the central maxima in Young's double slit experimeny is ${\mathrm{I}}_{0}$. Find out the intensity at a point where the path difference is

(a)                                                            (1/2) marks

Path difference =${\mathrm{S}}_{2}\mathrm{P}-{\mathrm{S}}_{1}\mathrm{P}$

Now $\left({\mathrm{S}}_{2}\mathrm{P}{\right)}^{2}-\left({\mathrm{S}}_{1}\mathrm{P}{\right)}^{2}=\left[{\mathrm{D}}^{2}+{\left(\mathrm{x}+\frac{\mathrm{d}}{2}\right)}^{2}\right]-\left[{\mathrm{D}}^{2}+{\left(\mathrm{x}+\frac{\mathrm{d}}{2}\right)}^{2}\right]$

=2xd

where${\mathrm{S}}_{1}{\mathrm{S}}_{2}$=d and OP=x

$\therefore {\mathrm{S}}_{2}\mathrm{P}-{\mathrm{S}}_{1}\mathrm{P}=\frac{2\mathrm{xd}}{\left({\mathrm{S}}_{2}\mathrm{P}+{\mathrm{S}}_{1}\mathrm{P}\right)}$

For x<<D and d<<D, we can write                                                   (1/2) marks

${\mathrm{S}}_{2}\mathrm{P}+{\mathrm{S}}_{1}\mathrm{P}\cong 2\mathrm{D}$                                                                               (1/2) marks

Hence, Path difference = ${\mathrm{S}}_{2}\mathrm{P}-{\mathrm{S}}_{1}\mathrm{P}=\frac{2\mathrm{xd}}{2\mathrm{D}}=\frac{\mathrm{xd}}{\mathrm{D}}$                                (1/2) marks

For constructive interference, we must have

$\frac{\mathrm{xd}}{\mathrm{D}}=\mathrm{n\lambda }$                                                                                         (1/2) marks

(1/2) marks

For destructive interference , we must have

$\frac{\mathrm{xd}}{\mathrm{D}}=\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{\lambda }$                                                                                (1/2) marks

(b) The general expression, for the intensity, at a point is $\mathrm{I}={\mathrm{I}}_{0}{\mathrm{cos}}^{2}\mathrm{\varphi }}{2}$

(i) For path difference = $\mathrm{\lambda }}{6}$            $\mathrm{\varphi }=60°$                               (1/2) marks

$\mathrm{I}=3{\mathrm{I}}_{0}}{4}$

(ii) For path difference = $\mathrm{\lambda }}{4}$           $\mathrm{\varphi }=90°$                                (1/2) marks

$\mathrm{I}={\mathrm{I}}_{0}}{2}$

(iii) For path difference = $\mathrm{\lambda }}{6}$           $\mathrm{\varphi }=120°$                              (1/2) marks

$\mathrm{I}={\mathrm{I}}_{0}}{4}$

Difficulty Level:

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