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Byju's Answer

Standard XII

Physics

YDSE Vs Single Slit Diffraction

The fringe wi...

Question

The fringe width $β$ of the fringes produced in Young's double slit experiment conducted for monochromatic light of wavelength $λ$ with slit difference $d$ and separation between slits and screen $D$ is :

β = \frac{λ D}{d}

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β = \frac{D d}{λ}

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β = \frac{λ d}{D}

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β = \frac{λ d^{2}}{D^{2}}

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Solution

The correct option is A $β = \frac{λ D}{d}$
From path difference
$Δ x = P S_{2} - P S_{1} \approx P S_{2} - P A$
$= S_{2} A = d s i n θ$
$\approx d t a n θ = \frac{d y}{D}$

fringe width $β = \frac{D λ}{d}$

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The fringe width β of the fringes produced in Young's double slit experiment conducted for monochromatic light of wavelength λ with slit difference d and separation between slits and screen D is :

The correct option is A β=λDdFrom path differenceΔx=PS2−PS1≈PS2−PA=S2A=dsinθ≈dtanθ=dyDfringe width β=Dλd

The fringe width $β$ of the fringes produced in Young's double slit experiment conducted for monochromatic light of wavelength $λ$ with slit difference $d$ and separation between slits and screen $D$ is :

The correct option is A $β = \frac{λ D}{d}$
From path difference
$Δ x = P S_{2} - P S_{1} \approx P S_{2} - P A$
$= S_{2} A = d s i n θ$
$\approx d t a n θ = \frac{d y}{D}$

fringe width $β = \frac{D λ}{d}$