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

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Quizplus · Accepted Answer

The distance between adjacent bright fringes is given by:$$d\sin	heta=m\lambda$$where d is the slit separation, θ is the angle of the bright fringe from the central fringe, m is the order of the bright fringe (m = 0 for the central fringe, m = 1 for the first bright fringe, etc.), and λ is the wavelength of the light.In this case, d = 3.00 × 10^-5 m, θ = tan^-1(10.0 cm / 2.00 m) = 2.86°, m = 7, and λ is unknown.Solving for λ, we get:$$\lambda=\frac{d\sin	heta}{m}=\frac{(3.00	imes10^{-5}	ext{ m})(\sin2.86^\circ)}{7}=214	ext{ nm}$$Therefore, the wavelength of the light is 214 nm.

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