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 7th bright fringe on the screen is 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

Question

Quizplus · Accepted Answer

The distance between adjacent bright fringes is given by:$$d\sin\theta=m\lambda$$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:$$\lambda=\frac{d\sin\theta}{m}=\frac{(3.00\times10^{-5}\text{ m})(\sin2.86^\circ)}{7}=214\text{ nm}$$Therefore, the wavelength of the light is 214 nm.

In a Two-Slit Experiment, the Slit Separation Is 3 10−510 ^ { - 5 }10−5

In a Two-Slit Experiment, the Slit Separation Is 3 $10 ^ { - 5 }$