The angular width of the central maximum in a single slit diffraction pattern is $60 ° .$ The width of the slit is $1 μ m .$ The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1cm, what is slit separation distance?
(i.e. distance between the centre of each slit)

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50 $μ m$

100 $μ m$

75 $μ m$

25 $μ m$

answer is A.

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Detailed Solution

Angular position of first minima
$b \sin θ = λ$ $\Rightarrow \sin θ = \frac{λ}{b}$

So wavelength of light used
$λ = b \sin λ = b \sin 30 ° = \frac{b}{2}$
Given $2 θ = 60 °$
$\Rightarrow θ = 30 °$
If similar two slits are used in YDSE at a separation ‘d’ then we have
Question Image
Fringe width in YDSE interference pattern is given as
$β = \frac{λ D}{d} = \frac{b D}{2 d}$

$So, slit separation is$

$d = \frac{b D}{2 β} = \frac{10^{- 6} \times 0.5}{2 \times 0.01} = 25 \times 10^{- 6} m$ $\Rightarrow d = 25 μ m$

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