A Young's double-slit apparatus is set up where a screen is positioned 0.80 m from the double slits.If the distance between alternating bright fringes is 0.83 cm,and the light source has a wavelength of 580 nm,what is the separation of the double slits? (1 nm = 10−⁹ m) A) 6.0 × 10−⁵ m B) 2.8 × 10−⁵ m C) 4.9 × 10−⁵ m D) 5.6 × 10−⁵ m

Question

Quizplus · Accepted Answer

The separation of the double slits (d) can be calculated using the formula for fringe spacing in Young's double-slit experiment: $ \Delta y = \frac{\lambda L}{d} $. Rearranging for d gives $ d = \frac{\lambda L}{\Delta y} $. Substituting the given values: $ \lambda = 580 \times 10^{-9} $ m, $ L = 0.80 $ m, and $ \Delta y = 0.0083 $ m, we find $ d = \frac{580 \times 10^{-9} \times 0.80}{0.0083} \approx 5.6 \times 10^{-5} $ m.

A Young's Double-Slit Apparatus Is Set Up Where a Screen