A single-slit diffraction pattern is displayed on a screen 0.900 m away from the slit.If the wavelength of the light is 600 nm and the slit is 1.50 $\times$ 10^-4 m wide,the distance from the first minimum on the right to the first minimum on the left is A)0.164 mm B)1.83 mm C)3.99 mm D)7.20 mm E)7.98 mm

Question

Quizplus · Accepted Answer

The distance between the first minimum on the right and the first minimum on the left is given by:$$d = \frac{2\lambda L}{w}$$where:* d is the distance between the first minimum on the right and the first minimum on the left* $\lambda$ is the wavelength of the light* L is the distance from the slit to the screen* w is the width of the slitPlugging in the given values, we get:$$d = \frac{2(600 	imes 10^{-9} 	ext{ m})(0.900 	ext{ m})}{1.50 	imes 10^{-4} 	ext{ m}} = 0.164 	ext{ mm}$$Therefore, the correct answer is A.

A Single-Slit Diffraction Pattern Is Displayed on a Screen 0 ×\times×

A Single-Slit Diffraction Pattern Is Displayed on a Screen 0 $\times$