In a Young's double-slit experiment, a fifth-order maximum occurs at an angle of 1.422°. If the screen is 3.90 m from the slits and the slit separation is 0.135 mm, what wavelength is being used?
A) 335 nm
B) 670 nm
C) 476 nm
D) 589 nm
E) 690 nm

Question

Quizplus · Accepted Answer

The formula for the position of maxima in a double-slit experiment is $d \sin \theta = m \lambda$, where $d$ is the slit separation, $\theta$ is the angle, $m$ is the order of the maximum, and $\lambda$ is the wavelength. Substituting the given values: $0.135 \times 10^{-3} \sin(1.422^\circ) = 5 \lambda$. Solving for $\lambda$ gives approximately 670 nm.

In a Young's Double-Slit Experiment, a Fifth-Order Maximum Occurs at an Angle