Light from a monochromatic source shines through a double slit onto a screen 5.00 m away.The slits are 0.180 mm apart.The dark bands on the screen are measured to be 1.70 cm apart.What is the wavelength of the incident light?
A) 457 nm
B) 306 nm
C) 392 nm
D) 612 nm
E) 784 nm

Question

Quizplus · Accepted Answer

The distance between dark bands (fringe spacing) is given by the formula: $ \Delta y = \frac{\lambda L}{d} $, where $ \Delta y $ is the fringe spacing, $ \lambda $ is the wavelength, $ L $ is the distance to the screen, and $ d $ is the slit separation. Rearranging for $ \lambda $, we have $ \lambda = \frac{\Delta y \cdot d}{L} $. Substituting the given values: $ \Delta y = 0.017 $ m, $ d = 0.00018 $ m, and $ L = 5.00 $ m, we find $ \lambda = \frac{0.017 \times 0.00018}{5.00} = 6.12 \times 10^{-7} $ m, or 612 nm.

Light from a Monochromatic Source Shines Through a Double Slit