Double-slit interference: A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe?
A) 23.9 cm
B) 5.96 cm
C) 5.59 cm
D) 2.63 cm

Question

Quizplus · Accepted Answer

The distance between adjacent bright fringes is given by:$$d = \frac{\lambda L}{d}$$where:* d is the distance between the slits* L is the distance between the slits and the screen* lambda is the wavelength of the lightPlugging in the given values, we get:$$d = \frac{(588 	imes 10^{-9} 	ext{ m})(0.11 	ext{ m})}{2464 	imes 10^{-9} 	ext{ m}} = 2.63 	imes 10^{-2} 	ext{ m} = 2.63 	ext{ cm}$$Therefore, the distance between the third and fourth bright fringes away from the central fringe is:$$3d = 3(2.63 	ext{ cm}) = 7.89 	ext{ cm}$$

Double-Slit Interference: a Double Slit Illuminated with Light of Wavelength