A double slit that is illuminated with coherent light of wavelength $644 \mathrm {~nm}$ produces a pattern of bright and dark fringes on a screen $6.00 \mathrm {~cm}$ from the slits. If the slits are $2783 \mathrm {~nm}$ apart, what is the distance on the screen between the $4 ^ { \text {th } }$ and the $2 ^ { \text {nd } }$ bright fringes on one side of the central maximum?
A) $3.13 \mathrm {~cm}$
B) $11.5 \mathrm {~cm}$
C) $17.6 \mathrm {~cm}$
D) $14.7 \mathrm {~cm}$
E) $23.0 \mathrm {~cm}$

Question

Quizplus · Accepted Answer

The distance between fringes in a double slit experiment is given by $y = \frac{m\lambda L}{d}$, where $y$ is the distance from the central maximum, $m$ is the fringe number, $\lambda$ is the wavelength of the light, $L$ is the distance to the screen, and $d$ is the distance between the slits. To find the distance between the 4th and 2nd bright fringes, calculate the positions of both and subtract. For the 4th fringe ($m=4$): $y_4 = \frac{4 \times 644 \times 10^{-9} \times 0.06}{2783 \times 10^{-9}}$. For the 2nd fringe ($m=2$): $y_2 = \frac{2 \times 644 \times 10^{-9} \times 0.06}{2783 \times 10^{-9}}$. Subtracting $y_2$ from $y_4$ gives the distance between these fringes, which is approximately $3.13 \mathrm {~cm}$, making option A correct.

A Double Slit That Is Illuminated with Coherent Light of Wavelength