Helium-neon laser light (λ = 632.8 nm)is sent through a single slit of width 0.55 mm.What is the width of the central maximum on a screen 1.4 m in back of the slit? (1 nm = 10−⁹ m) A)1.6E-3 m B)8.0E-4 m C)3.2E-3 m D)7.5E-5 m E)1.5E-4 m

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Quizplus · Accepted Answer

The width of the central maximum ($W$) in a single-slit diffraction pattern is given by the formula $W = 2L\lambda / a$, where $L$ is the distance from the slit to the screen (1.4 m), $\lambda$ is the wavelength of the light (632.8 nm = $632.8 \times 10^{-9}$ m), and $a$ is the width of the slit (0.55 mm = $0.55 \times 10^{-3}$ m). Substituting the given values, $W = 2 \times 1.4 \times (632.8 \times 10^{-9}) / (0.55 \times 10^{-3}) = 3.2 \times 10^{-3}$ m or 3.2E-3 m.

Helium-Neon Laser Light (λ = 632