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Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source $\left ( \lambda = 632.8\; nm \right )$ . The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at a distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to:
Option: 1 $1.27 \;\mu m$

Option: 2 $2.87 \;n m$
Option: 3 $2 \;n m$
Option: 4 $2.05 \;\mu m$

Answers (1)

best_answer

$\mathrm{y}=\frac{\mathrm{nD\lambda}}{\mathrm{d}}$

or

$\begin{aligned} &\mathrm{n}=\frac{\mathrm{yd}}{\mathrm{D} \lambda}=\frac{1.27 \times 10^{-3} \times 10^{-3}}{1 \times 632.8 \times 10^{-9}}=2\\ &\text {So Path difference } \Delta \mathrm{x}=\mathrm{n} \lambda\\ &\begin{array}{l} =2 \times 632.8 \mathrm{nm} \\ =1265.6 \mathrm{nm} \\ =1.27 \mu \mathrm{m} \end{array} \end{aligned}$

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