First slide

Young's double slit Experiment

Question

Two ideal slits $S_{1} and S_{2}$ are at a distance d apart, and illuminated by light of wavelength $λ$ passing through an ideal source slit S placed on the line through S₂ 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

Moderate

A

$\sqrt{\frac{3 λ D}{2}}$
B

$\sqrt{λ D}$
C

$\sqrt{\frac{λ D}{2}}$
D

$\sqrt{3 λ D}$

Solution

Path difference between the waves reaching at P,
$Δ = Δ_{1} + Δ_{2}$
where $Δ_{1}$ = Initial path difference
$Δ_{2} =$ Path difference between the waves after emerging from slits.
$Δ_{1} = {SS}_{1} - {SS}_{2} = \sqrt{D^{2} + d^{2}} - D$
$and Δ_{2} = S_{1} O - S_{2} O = \sqrt{D^{2} + d^{2}} - D$
$Δ = 2 \{{(D^{2} + d^{2})}^{\frac{1}{2}} - D\} = 2 \{(D^{2} + \frac{d^{2}}{2 D}) - D\}$
$= \frac{d^{2}}{D}$ (From Binomial expansion)
For obtaining dark at O, $Δ$ must be equals to
$(2 n - 1) \frac{λ}{2}, i.e., \frac{d^{2}}{D} = (2 n - 1) \frac{λ}{2} \Rightarrow d = \sqrt{\frac{(2 n - 1) λD}{2}}$
For minimum distance n = 1 so $d = \sqrt{\frac{λD}{2}} .$

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