Two ideal slits $S_{1}$ and $S_{2}$ are at a distance $d$ apart, and illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_{2}$ as shown. The distance between the planes of slits and the source slit is $D$. A screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is

Question

Dr. Abubacker · Accepted Answer

(c) Path difference between the waves reaching at $O,\ \mathrm{\Delta} = \mathrm{\Delta}_{1} + \mathrm{\Delta}_{2}$ where $\mathrm{\Delta}_{1} =$ Initial path difference $\mathrm{\Delta}_{2} =$ Path difference between the waves after emerging from slits
$\mathrm{\Delta}_{1} = SS_{1} - SS_{2} = \sqrt{D^{2} + d^{2}} - D$ and
$\mathrm{\Delta}_{2} = S_{1}O - S_{2}O = \sqrt{D^{2} + d^{2}} - D$
$	herefore\mathrm{\Delta} = 2\left\{ \left( D^{2} + d^{2} ight)^{\frac{1}{2}} - D ight\} = 2\left\{ \left( D + \frac{d^{2}}{2D} ight) - D ight\}$
$= \frac{d^{2}}{D}$ (From Binomial expansion)
For obtaining dark at $O,\mathrm{\Delta}$ must be equals to $(2n - 1)\frac{\lambda}{2}\ i.e\frac{d^{2}}{D} = (2n - 1)\frac{\lambda}{2} \Rightarrow \sqrt{\frac{(2n - 1)\lambda D}{2}} = d$
For minimum distance $n = 1$ so $d = \sqrt{\frac{\lambda D}{2}}$

Explanation

Reviewed by Dr. Abubacker

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