A two-slit arrangement with $60.3 \mu \mathrm { m }$ separation between the slits is illuminated with $491.0 \mathrm {~nm}$ coherent light. If a viewing screen is located $2.14 \mathrm {~m}$ from the slits, find the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side.
A) $34.9 \mathrm {~mm}$
B) $69.7 \mathrm {~mm}$
C) $24.6 \mathrm {~mm}$
D) $52.3 \mathrm {~mm}$

Question

Quizplus · Accepted Answer

The distance between dark fringes on either side of the central maximum can be calculated using the formula for the position of dark fringes in a two-slit interference pattern: $y = \frac{(m + 1/2)\lambda L}{d}$, where $y$ is the distance from the central maximum to the $m$th dark fringe, $\lambda$ is the wavelength of the light, $L$ is the distance from the slits to the screen, and $d$ is the separation between the slits. For the first dark fringe on one side ($m = 0$) and the second dark fringe on the other side ($m = 1$), the total distance between these fringes is $2y$, where $y$ for $m = 0$ is calculated with $m + 1/2 = 1/2$, and for $m = 1$, $m + 1/2 = 3/2$. However, since we want the distance from the first dark fringe on one side to the second on the other, we calculate the positions of these fringes relative to the central maximum and add them.For the first dark fringe ($m = 0$): $y_1 = \frac{(1/2) \cdot 491.0 \times 10^{-9} \cdot 2.14}{60.3 \times 10^{-6}}$.For the second dark fringe on the opposite side ($m = 1$): $y_2 = \frac{(3/2) \cdot 491.0 \times 10^{-9} \cdot 2.14}{60.3 \times 10^{-6}}$.The total distance between these fringes is simply $y_2 - y_1$, which, due to symmetry about the central maximum, is equivalent to calculating $2y_1$ (since $y_2$ is exactly three times $y_1$, and the difference between three parts and one part is two parts).Thus, $2y_1 = 2 \cdot \frac{(1/2) \cdot 491.0 \times 10^{-9} \cdot 2.14}{60.3 \times 10^{-6}} = \frac{491.0 \times 10^{-9} \cdot 2.14}{60.3 \times 10^{-6}}$.Plugging in the numbers: $2y_1 = \frac{491.0 \times 10^{-9} \cdot 2.14}{60.3 \times 10^{-6}} \approx 34.9 \times 10^{-3} \mathrm{m} = 34.9 \mathrm{mm}$.Therefore, the correct answer is $34.9 \mathrm{mm}$, which corresponds to choice A.

A Two-Slit Arrangement With 60.3μm60.3 \mu \mathrm { m }60.3μm Separation Between the Slits Is Illuminated With

A Two-Slit Arrangement With $60.3 \mu \mathrm { m }$ Separation Between the Slits Is Illuminated With