A) \[\lambda =\frac{{{b}^{2}}}{d}\]
B) \[\lambda =\frac{2{{b}^{2}}}{d}\]
C) \[\lambda =\frac{{{b}^{2}}}{3d}\]
D) \[\lambda =\frac{2{{b}^{2}}}{3d}\]
Correct Answer: A
Solution :
Path difference between the rays reaching infront of slit S1 is. \[{{S}_{1}}P-{{S}_{2}}P={{({{b}^{2}}+{{d}^{2}})}^{1/2}}-d\] For distructive interference at P \[{{S}_{1}}P-{{S}_{2}}P=\frac{(2\,n-1)\lambda }{2}\] i.e., \[{{({{b}^{2}}+{{d}^{2}})}^{1/2}}-d=\frac{(2n-1)\lambda }{2}\] \[\Rightarrow d\,{{\left( 1+\frac{{{b}^{2}}}{{{d}^{2}}} \right)}^{1/2}}-d=\frac{(2n-1)\lambda }{2}\] \[\Rightarrow d\,\left( 1+\frac{{{b}^{2}}}{2{{d}^{2}}}+...... \right)-d=\frac{(2n-1)\lambda }{2}\] (Binomial Expansion) \[\Rightarrow \frac{b}{2d}=\frac{(2n-1)\lambda }{2}\Rightarrow \lambda =\frac{{{b}^{2}}}{(2n-1)d}\] For \[n=1,\ 2............,\ \lambda =\frac{{{b}^{2}}}{d},\ \frac{{{b}^{2}}}{3d}\]You need to login to perform this action.
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