Coherent monochromatic light illuminates two narrow slits separated by 2.400 × 10^-4 m. A screen is 1.9 m away. The distance from the second bright fringe on one side of the central maximum to the sixth bright fringe on the other side is 0.03103333 m. Assume small angles and negligible slit-width envelope effects. Determine the wavelength. Predict the fringe spacing if screen distance increases by 50%, and explain why measuring this multi-fringe span is preferable to measuring one spacing.

Question

Accepted Answer

Wavelength 4.900 × 10^-7 m; new fringe spacing 0.005819 m. The measured span contains eight fringe intervals. Assign fringe orders −2 and +6 to the two endpoints. Their separation contains 6 − (−2) = 8 intervals, not four or nine; the central maximum is an order-zero location. Therefore one fringe spacing is s = measured span/8 = 0.003879 m. For Young's interference at small angles, s = λD/d. Rearranging gives λ = sd/D = 4.900 × 10^-7 m. This lies in the visible-light range, providing a useful check on the metre and millimetre conversions. With wavelength and slit separation unchanged, spacing is proportional to screen distance. Increasing D by 50% gives s_new = 1.5s = 0.005819 m. Locating two endpoints over a long span usually gives similar absolute position uncertainty to measuring one spacing, but a much larger measured distance. Dividing by eight then reduces fractional spacing uncertainty; repeat readings and avoid parallax when using a ruler.

Measure wavelength using multiple fringes · set 4

What a complete response should do

How did this one feel?