question_answer

Two ideal slits S1 and S2 are at a distance d apart, and illuminated by light of wavelength l passing through an ideal source slit S placed on the line through S2 as shown. The distance between the planes of slits and the source slit is D. A screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is

A)            \[\sqrt{\frac{3\lambda D}{2}}\]

B)            \[\sqrt{\lambda D}\]

C)            \[\sqrt{\frac{\lambda D}{2}}\]

D)            \[\sqrt{3\lambda D}\]

Correct Answer: C

Solution :

           Path difference between the waves reaching at\[P,\]\[\Delta ={{\Delta }_{1}}+{{\Delta }_{2}}\] where\[{{\Delta }_{1}}=\] Initial path difference \[{{\Delta }_{2}}=\]Path difference between the waves after emerging from slits. \[{{\Delta }_{1}}=S\,{{S}_{1}}-S\,{{S}_{2}}=\sqrt{{{D}^{2}}+{{d}^{2}}}-D\] and \[{{\Delta }_{2}}={{S}_{1}}O-{{S}_{2}}O=\sqrt{{{D}^{2}}+{{d}^{2}}}-D\] \[\therefore \,\,\,\Delta =2\left\{ {{({{D}^{2}}+{{d}^{2}})}^{\frac{1}{2}}}-D \right\}=2\left\{ ({{D}^{2}}+\frac{{{d}^{2}}}{2D})-D \right\}\]         \[=\frac{{{d}^{2}}}{D}\]       (From Binomial expansion) For obtaining dark at O, \[\Delta \] must be equals to \[(2n-1)\frac{\lambda }{2}\]  i.e. \[\frac{{{d}^{2}}}{D}=(2n-1)\frac{\lambda }{2}\Rightarrow d\sqrt{\frac{(2n-1)\lambda \,D}{2}}\] For minimum distance \[n=1\] so \[d=\sqrt{\frac{\lambda \,D}{2}}\]