In Young's double slit experiment, the light has a frequency of $6 × 10^{14} Hz$ and the distance between the centers of adjacent fringes is 0.75 mm. If the screen is 1.5 m away, what is the distance between the slits?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $10^{-3} m$
In Young's Double Slit Experiment,
$Δy=\frac{λL}{d}$
and,
Wavelength, $λ=\frac{c}{f}=\frac{3×10^8}{6×10^{14}}=5×10^{-7}m$
Also,
$d=\frac{λL}{Δy}=\frac{5×10^{-7}×1.5}{0.75×10^{-3}}$
$=1mm=10^{-3} m$