Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
In Young's double slit experiment, white light is used. The separation between the slits is b. The screen is at a distance\[d(d>>b)\]from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are (1) \[\frac{{{b}^{2}}}{d}\] (2) \[\frac{2{{b}^{2}}}{d}\] (3) \[\frac{{{b}^{3}}}{3d}\] (4) \[\frac{2{{b}^{2}}}{3d}\]A) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: D
Solution :
Path difference between the rays reaching in front of slit\[{{S}_{1}}\]is \[{{S}_{1}}P-{{S}_{2}}P={{({{b}^{2}}+{{d}^{2}})}^{1/2}}-d\] For destructive interference at P \[{{S}_{1}}P-{{S}_{2}}P=\frac{(2n-1)}{2}\lambda \] ie, \[{{({{b}^{2}}+{{d}^{2}})}^{1/2}}-d=\frac{(2n-1)\lambda }{2}\] \[\Rightarrow \]\[d{{\left( 1+\frac{{{b}^{2}}}{{{d}^{2}}} \right)}^{1/2}}-d=\frac{(2n-1)\lambda }{2}\] \[\Rightarrow \]\[d\left[ 1+\frac{{{b}^{2}}}{2{{d}^{2}}}+..... \right]-d=\frac{(2n-1)\lambda }{2}\] Binomial expansion \[\Rightarrow \] \[\frac{b}{2d}=\frac{(2n-1)\lambda }{2}\] \[\Rightarrow \] \[\lambda =\frac{{{b}^{2}}}{(2n-1)d}\] For \[n=1,2....\] \[\lambda =\frac{{{b}^{2}}}{d},\frac{{{b}^{2}}}{3d}\]
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