Physics

Physics

Problem

A single slit of fixed, unknown width is illuminated by monochromatic light. The resulting diffraction pattern is projected onto two separate screens placed normal to the beam at distances L1=1.0 mL_1 = 1.0\,\text{m} and L2=3.0 mL_2 = 3.0\,\text{m} from the slit. For each screen, the experimenter measures the average spacing Δy\Delta y between successive dark bands. The data are summarized: Screen 1: Δy1=4.0 mm\Delta y_1 = 4.0\,\text{mm} Screen 2: Δy2=11.9 mm\Delta y_2 = 11.9\,\text{mm} (a) Derive an expression for the spacing Δy\Delta y between adjacent dark bands on a screen in terms of the unknown slit width aa, the wavelength λ\lambda, and the screen distance LL. State any approximations you introduce. (b) Using the expression from part (a), obtain the theoretical ratio Δy2Δy1\dfrac{\Delta y_2}{\Delta y_1} expected for the two screen distances given. (c) Compare the theoretical ratio with the experimental value determined from the data. Evaluate whether the data are consistent with theory within reasonable experimental uncertainty. (d) Identify one plausible systematic source of error in the measurement of Δy\Delta y and discuss qualitatively how it could affect the comparison performed in part (c).
Intuitive explanation

Single-slit diffraction dark fringes are spaced proportionally to the distance from the slit: Δy=λLa\Delta y = \frac{\lambda L}{a}, where aa is the slit width. Since you have two measurements at different distances, you can set up a ratio to eliminate the unknown wavelength and solve for aa—then use either dataset to find λ\lambda. The key is recognizing that the ratio of spacings directly reveals the slit width without needing to know the light's wavelength beforehand.

Intuitive explanation

Single-slit diffraction dark fringes are spaced proportionally to the distance from the slit: Δy=λLa\Delta y = \frac{\lambda L}{a}, where aa is the slit width. Since you have two measurements at different distances, you can set up a ratio to eliminate the unknown wavelength and solve for aa—then use either dataset to find λ\lambda. The key is recognizing that the ratio of spacings directly reveals the slit width without needing to know the light's wavelength beforehand.

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