ACT Science Practice Question #1855 (Hard (560)) | Test Citadel
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ACT Science Difficulty: Hard (560)

Digital ACT Science Practice Question #1855

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If the red laser (632.8 nm) were replaced with a violet laser (405.0 nm) at d = 0.20 mm and L = 2.0 m, the fringe spacing would: Experiment Context (Interference Fringe Spacing of Laser Light): Students directed a 632.8 nm helium-neon red laser through a double-slit aperture onto a screen 2.0 m away, measuring fringe spacing (y) between adjacent bright fringes: Table 34: - Slit Separation d = 0.10 mm: Screen Distance L = 2.0 m; Fringe Spacing y = 12.7 mm - Slit Separation d = 0.20 mm: Screen Distance L = 2.0 m; Fringe Spacing y = 6.3 mm - Slit Separation d = 0.40 mm: Screen Distance L = 2.0 m; Fringe Spacing y = 3.2 mm - Slit Separation d = 0.20 mm: Screen Distance L = 4.0 m; Fringe Spacing y = 12.7 mm
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Shorter wavelength = narrower fringe spacing.

Masterclass Solution & Distractor Trap Analysis

y is directly proportional to wavelength lambda (y = lambda * L / d). Smaller lambda yields smaller y....

Distractor Analysis: Trap choice eliminates careless test-takers who confuse roots with coordinates...

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