Question

In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890$ $\mathring{A}$ is :-

• A. $1178 \times 10^{-9} m$
• B. $1178 \times 10^{-6} m$
• C. $1178 \times 10^{-12} m$
• D. $5890 \times 10^{-7} m$

Answer 1

Answer: (B)

$\beta=\frac{\lambda D}{d}=\frac{5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}}$
$=589 \times 10^{-6} m$
Distance between first and third bright fringe is $2 \beta=2 \times 589 \times 10^{-6} m$
$=1178 \times 10^{-6} m $