In a two-slit experiment, the slit separation is $3.00 \times 10 ^ { - 5 } \mathrm {~m}$. The interference pattern is created on a screen that is $2.00 \mathrm {~m}$ away from the slits. If the 7 th bright fringe on the screen is $10.0 \mathrm {~cm}$ away from the central fringe, what is the wavelength of the light?
A) $224 \mathrm {~nm}$
B) $214 \mathrm {~nm}$
C) $204 \mathrm {~nm}$
D) $234 \mathrm {~nm}$
E) $100 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The wavelength of the light can be calculated using the formula for the position of the bright fringes in a two-slit interference pattern: $y = \frac{m\lambda D}{d}$, where $y$ is the distance from the central maximum to the m-th bright fringe, $m$ is the order number of the fringe, $\lambda$ is the wavelength of the light, $D$ is the distance from the slits to the screen, and $d$ is the slit separation. Rearranging for $\lambda$, we get $\lambda = \frac{yd}{mD}$. Substituting the given values ($y = 0.1 \, \text{m}$, $d = 3.00 \times 10^{-5} \, \text{m}$, $m = 7$, $D = 2.00 \, \text{m}$), we find $\lambda = \frac{0.1 \times 3.00 \times 10^{-5}}{7 \times 2.00} = 214 \, \text{nm}$.

In a Two-Slit Experiment, the Slit Separation Is 3.00×10−5 m3.00 \times 10 ^ { - 5 } \mathrm {~m}3.00×10−5 m

In a Two-Slit Experiment, the Slit Separation Is $3.00 \times 10 ^ { - 5 } \mathrm {~m}$