In a Young's double slit experiment,a 5^th order maximum occurs at an angle of 1.422$^{\circ}$.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 wavelength can be calculated using the formula for the position of maxima in a double slit experiment: $d \sin \theta = m \lambda$. Here, $d = 0.135 \, \text{mm} = 0.135 \times 10^{-3} \, \text{m}$, $\theta = 1.422^\circ$, and $m = 5$. Solving for $\lambda$, we get $\lambda = \frac{d \sin \theta}{m} = \frac{0.135 \times 10^{-3} \times \sin(1.422^\circ)}{5} \approx 670 \, \text{nm}$.

In a Young's Double Slit Experiment,a 5th Order Maximum Occurs ∘^{\circ}∘

In a Young's Double Slit Experiment,a 5^th Order Maximum Occurs $^{\circ}$