Question:
In Young's double slit arrangement, slits are separated by a gap of $0.5 \mathrm{~mm}$, and the screen is placed at a distance of $0.5 \mathrm{~m}$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890A is :-
Correct Option: 2,
Solution:
$\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}}$
$=589 \times 10^{-6} \mathrm{~m}$
Distance between first and third bright fringe
is $2 \beta=2 \times 589 \times 10^{-6} \mathrm{~m}$
$=1178 \times 10^{-6} \mathrm{~m}$
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