In a two-slit experiment, the slit separation is 3.00 × 10^-5m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? A) 100 nm B) 204 nm C) 214 nm D) 224 nm E) 234 nm

Question

Quizplus · Accepted Answer

The distance between the central fringe and the seventh bright fringe (n=7) is given by:

y = nλL/d

where y is the distance from the central fringe to the nth bright fringe, λ is the wavelength of the light, L is the distance from the slits to the detector, and d is the slit spacing.

Substituting the given values, we have:

10.0 cm = 7λ(2.00 m)/(3.00 × 10^-5 m)
λ = (10.0 cm)(3.00 × 10^-5 m)/(7 × 2.00 m) = 2.14 × 10^-7 m = 214 nm

Therefore, the wavelength of the light passing through the slits is 214 nm, which corresponds to choice C.

In a Two-Slit Experiment, the Slit Separation Is 3