Question Detail
Question
Light of wavelength 580 nm is incident on a slit of width 0.300 mm. An observing screen is placed 2.00m from the slit. Find the position of the first order dark fringe from the center of the screen.
Correct Answer
Option B is the correct answer.
Detailed Explanation
This question likely refers to the first minimum (dark fringe) in a double-slit interference pattern, where the 'slit width' is actually the 'slit separation' (d). For destructive interference (dark fringes) in a double-slit experiment, the path difference is (m + 0.5)λ. The position (y) on the screen is given by y = (m + 0.5)λL/d, where m is the order of the fringe (m=0 for the first dark fringe).Given:Wavelength (λ) = 580 nm = 580 × 10^-9 mSlit separation (d) = 0.300 mm = 0.300 × 10^-3 mDistance to screen (L) = 2.00 mFor the first order dark fringe, m = 0.y = (0 + 0.5) × (580 × 10^-9 m) × (2.00 m) / (0.300 × 10^-3 m)y = 0.5 × 580 × 2 × 10^(-9 + 3) / 0.3 my = 580 × 10^-6 / 0.3 my = 1933.33 × 10^-6 my ≈ 1.933 mmThis value is closest to option B.
Hint
For destructive interference (dark fringes) in a double-slit experiment, the path difference is an odd multiple of half-wavelengths, and the position is given by y = (m + 0.5)λL/d.