Derive the formula $ω = \frac{D λ}{2}$ for fringe width in Young's double slit experiment. The symbols used have their usual meanings.

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Solution

$a sin θ =$ path difference $= λ$
$sin θ x θ$ for small angles
$θ = \frac{n λ}{a}$
$\frac{y}{D} = tan θ \equiv θ$
$y = D θ = \frac{n λ D}{d}$
$y$ is position of maxima
fring width = difference between maxima $= \frac{λ D}{d}$

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Derive the formula ω=Dλ2 for fringe width in Young's double slit experiment. The symbols used have their usual meanings.

asinθ= path difference =λsinθxθ for small angles θ=nλayD=tanθ≡θy=Dθ=nλDdy is position of maximafring width = difference between maxima =λDd

Derive the formula $ω = \frac{D λ}{2}$ for fringe width in Young's double slit experiment. The symbols used have their usual meanings.

$a sin θ =$ path difference $= λ$
$sin θ x θ$ for small angles
$θ = \frac{n λ}{a}$
$\frac{y}{D} = tan θ \equiv θ$
$y = D θ = \frac{n λ D}{d}$
$y$ is position of maxima
fring width = difference between maxima $= \frac{λ D}{d}$