In a single-slit experiment, light of wavelength 500 nm is used to produce a diffraction pattern on a screen 4.0 m away. If the distance between both third-order minima is 6.0 cm, what is the slit width? A) 2.0 × 10^-4 m B) 1.0 × 10^-4 m C) 4.0 × 10^-4 m D) 0.67 × 10^-4 m E) 0.44 × 10^-4 m

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Quizplus · Accepted Answer

The formula for the position of the nth minimum in a single-slit diffraction pattern is given by $y_n = \frac{n\lambda L}{a}$, where $y_n$ is the distance from the central maximum to the nth minimum, $\lambda$ is the wavelength of the light, $L$ is the distance from the slit to the screen, and $a$ is the slit width. For the third-order minimum ($n = 3$), the distance between both third-order minima is twice $y_3$, so $2y_3 = 6.0 \, \text{cm} = 0.06 \, \text{m}$. Substituting the given values ($\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}$, $L = 4.0 \, \text{m}$, and $2y_3 = 0.06 \, \text{m}$) into the rearranged formula $a = \frac{n\lambda L}{y_n}$ gives $a = \frac{3 \times 500 \times 10^{-9} \, \text{m} \times 4.0 \, \text{m}}{0.03 \, \text{m}} = 2.0 \times 10^{-4} \, \text{m}$.

In a Single-Slit Experiment, Light of Wavelength 500 Nm Is