Interference fringes are observed on a screen by illuminating two thin slits $1 \mathrm{~mm}$ apart with a light source $(\lambda$ $=632.8 \mathrm{~nm}$ ). The distance between the screen and the slits is $100 \mathrm{~cm}$. If a bright fringe is observed on a screen at a distance of $1.27 \mathrm{~mm}$ from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
Correct Option: 1
(1) Path difference, $\Delta P=d \sin \theta=d \theta$
$d=$ distance between slits $=1 \mathrm{~mm}=10^{-3} \mathrm{~mm}$
$D=$ distance between the slits and screen $=100 \mathrm{~cm}=1 \mathrm{~m}$
$y=$ distance between central bright fringe and observed
fringe $=1.27 \mathrm{~mm}$
$\therefore \Delta P=\frac{d y}{D}=\frac{10^{-3} \times 1.270 \mathrm{~mm}}{1 \mathrm{~m}}=1.27 \mu \mathrm{m}$
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