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Light Waves — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsLight Waves5 Marks
Question
Consider the arrangement shown in figure (17-E4). The distance D is large compared to the separation d between the slits.
  1. Find the minimum value of d so that there is a dark fringe at 0.
  2. Suppose d has this value. Find the distance x at which the next bright fringe is formed.
  3. Find the fringe-width.
✓

Answer

From the diagram, it can be seen that at point O. Path difference = (AB + BO) - (AC + CO) = 2(AB - AC) [Since, AB = BO and AC = CO] $=2\Big(\sqrt{\text{d}^2+\text{D}^2}-\text{D}\Big)$ For dark fringe, path difference should be odd multiple of $\frac{\lambda}{2}.$ So, $2\Big(\sqrt{\text{d}^2+\text{D}^2}-\text{D}\Big)=(2\text{n}+1)\Big(\frac{\lambda}{2}\Big)$$\Rightarrow\sqrt{\text{d}^2+\text{D}^2}=\text{D}+(2\text{n}+1)\Big(\frac{\lambda}{4}\Big)$
$\Rightarrow\text{D}^2+\text{d}^2=\text{D}^2+(2\text{n}+1)^2\frac{\lambda^2}{16}+(2\text{n}+1)\frac{\lambda\text{D}}{2}$
Neglecting, $(2\text{n}+1)^2\frac{\lambda^2}{16},$ as it is very small We get, $\text{d}=\sqrt{(2\text{n}+1)\frac{\lambda\text{D}}{2}}$ For minimum ‘d’, putting $\text{n}=0\Rightarrow\text{d}_\text{min}=\sqrt{\frac{\lambda\text{D}}{2}}.$

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