question_answer

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 [IIT 1984; AIIMS 1995]

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 \[n=1,\ 2............,\ \lambda =\frac{{{b}^{2}}}{d},\ \frac{{{b}^{2}}}{3d}\]