NEET-XII-Physics
17: Light Waves
- #26Consider 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
digAnsr: iAns :
From the figure, `` \,\mathrm{\,AB\,}=\,\mathrm{\,BO\,}\,\mathrm{\,and\,}\,\mathrm{\,AC\,}=\,\mathrm{\,CO\,}``.
Path difference of the wave front reaching O,
`` ∆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)``
For dark fringe to be formed at O, path difference should be an odd multiple of `` \frac{\lambda }{2}``.
`` \,\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}``
Neglecting, `` {\left(2n+1\right)}^{2}\frac{{\lambda }^{2}}{16}``
`` ``, as it is very small, we get:
`` d=\left(\sqrt{2n}+1\right)\frac{\lambda D}{2}``
`` \,\mathrm{\,For\,}\,\mathrm{\,minimum\,}d,\,\mathrm{\,putting\,},n=0``
`` {d}_{\,\mathrm{\,min\,}}=\sqrt{\left(\frac{\lambda D}{2}\right)}``, we get:
Thus, for `` {d}_{\,\mathrm{\,min\,}}=\sqrt{\left(\frac{\lambda D}{2}\right)}`` there is a dark fringe at O.
Page No 382: