Physics - Wave Optics Question with Solution | TestHub

Resolution limit$=\frac{\text{1.22}\mathrm{\lambda }}{\text{d}}$
Again resolution limit$=\text{sin}\mathrm{\theta }=\mathrm{\theta }=\frac{\mathrm{y}}{\text{D}}$
$\therefore$$\frac{\mathrm{y}}{\text{D}}=\frac{1·22\mathrm{\lambda }}{\text{d}}$


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$\text{D}=\frac{\mathrm{y}\text{d}}{\text{1.22}\mathrm{\lambda }}$

$\text{D}=\frac{\left({\text{10}}^{-3}\right)\times \left(3\times {\text{10}}^{-3}\right)}{\left(\text{1.22}\right)\times \left(5\times {\text{10}}^{-7}\right)}=\frac{\text{30}}{\text{6.1}}\approx 5\text{ m}$.