Question

Consider the arrangement shown in the figure (17-E4). The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe-width.

Figure

Answer 1

From the figure, $\mathrm{AB}=\mathrm{BO}\mathrm{and}\mathrm{AC}=\mathrm{CO}$.
Path difference of the wave front reaching O,
$$\begin{gathered} ∆x=\left(\mathrm{AB}+\mathrm{BO}\right)-\left(\mathrm{AC}+\mathrm{CO}\right) \\ =2\left(\mathrm{AB}-\mathrm{AC}\right) \\ =2\left(\sqrt{D^2+d^2}-D\right) \end{gathered}$$
For dark fringe to be formed at O, path difference should be an odd multiple of $\frac{\lambda }{2}$.
$$\begin{gathered} \mathrm{So},∆x=\left(2n+1\right)\frac{\lambda }{2} \\ \Rightarrow 2\left(\sqrt{D^2+d^2}-D\right)=\left(2n+1\right)\frac{\lambda }{2} \\ \Rightarrow \sqrt{D^2+d^2}=D+\left(2n+1\right)\frac{\lambda }{4} \\ \Rightarrow D^2+d^2=D^2+{\left(2n+1\right)}^2\frac{{\lambda }^2}{16}+\left(2n+1\right)\frac{\lambda D}{2} \end{gathered}$$
Neglecting, ${\left(2n+1\right)}^2\frac{{\lambda }^2}{16}$, as it is very small, we get:
$$\begin{gathered} d=\left(\sqrt{2n}+1\right)\frac{\lambda D}{2} \\ \mathrm{For}\mathrm{minimum}d,\mathrm{putting},n=0 \\ {d}_{\min }=\sqrt{\left(\frac{\lambda D}{2}\right)} \end{gathered}$$, we get:
Thus, for $d_{\min }=\sqrt{\left(\frac{\lambda D}{2}\right)}$ there is a dark fringe at O.