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

Question

Quizplus · Accepted Answer

The distance between the $m$th bright fringe and the central maximum in a double slit experiment is given by the formula $y_m = \frac{m\lambda L}{d}$, where $\lambda$ is the wavelength of the light, $L$ is the distance from the slits to the screen, $d$ is the distance between the slits, and $m$ is the order number of the fringe. To find the distance between the 4th and 2nd bright fringes, we calculate the positions of both and subtract them. For the 4th fringe ($m=4$): $y_4 = \frac{4 \times 644 \times 10^{-9} \times 0.06}{2783 \times 10^{-9}}$ meters. For the 2nd fringe ($m=2$): $y_2 = \frac{2 \times 644 \times 10^{-9} \times 0.06}{2783 \times 10^{-9}}$ meters. Subtracting $y_2$ from $y_4$, we find the distance between these fringes. Simplifying, $y_4 - y_2 = 2 \times 644 \times 10^{-9} \times 0.06 / 2783 \times 10^{-9}$ meters, which calculates to approximately 0.0313 meters or 3.13 cm.

A Double Slit That Is Illuminated with Coherent Light of Wavelength