A Young's interference experiment is performed with monochromatic green light ($\lambda$ = 470 nm).The separation between the slits is 0.120 mm,and the interference pattern on a screen shows the first side maximum 3.5 mm from the center of the pattern.How far away from the slits is the screen?

Question

Quizplus · Accepted Answer

The distance to the screen $D$ can be calculated using the formula for the position of the first side maximum in a double-slit interference pattern: $y = \frac{\lambda D}{d}$. Rearranging for $D$, we have $D = \frac{y \cdot d}{\lambda}$. Substituting $y = 3.5 \times 10^{-3}$ m, $d = 0.120 \times 10^{-3}$ m, and $\lambda = 470 \times 10^{-9}$ m, we find $D \approx 0.89$ m.

A Young's Interference Experiment Is Performed with Monochromatic Green Light λ\lambdaλ

A Young's Interference Experiment Is Performed with Monochromatic Green Light $\lambda$