Light of wavelength 418 nm is incident on a slit of width 0.35 mm,and a diffraction pattern is produced on a screen that is 2.4 m from the slit.What is the distance of the second dark fringe from the center of the bright fringe? (1 nm = 10−⁹ m) A)2.9E-7 m B)5.7E-3 m C)1.6E-3 m D)1.4E-6 m E)1.1E-6 m

Question

Quizplus · Accepted Answer

The distance of the nth dark fringe from the center in a single-slit diffraction pattern is given by $y_n = \frac{n\lambda L}{a}$, where $n$ is the fringe order (n=2 for the second dark fringe), $\lambda$ is the wavelength of the light (418 nm or $418 \times 10^{-9}$ m), $L$ is the distance from the slit to the screen (2.4 m), and $a$ is the width of the slit (0.35 mm or $0.35 \times 10^{-3}$ m). Substituting these values gives $y_2 = \frac{2 \times 418 \times 10^{-9} \times 2.4}{0.35 \times 10^{-3}} = 5.7 \times 10^{-3}$ m.

Light of Wavelength 418 Nm Is Incident on a Slit