Light of 600.0 nm is incident upon a single slit. The resulting diffraction pattern is observed on a screen that is 0.50 m from the slit. The distance between the first and third minima of the diffraction pattern is 0.80 mm. Which range of values listed below contains the width of the slit?
A)0.1 mm to 0.4 mm
B)0.4 mm to 0.8 mm
C)0.8 mm to 1.2 mm
D)1.2 mm to 1.6 mm
E)1.6 mm to 2.0 mm

Question

Quizplus · Accepted Answer

The distance between the first and third minima of the diffraction pattern is given by:$$d\sin	heta=3\lambda$$where d is the width of the slit, θ is the angle of diffraction, and λ is the wavelength of light.We can use the small angle approximation to get:$$d	heta\approx d	an	heta=\frac{d}{L}$$where L is the distance between the slit and the screen.Substituting this into the first equation, we get:$$d=\frac{3\lambda L}{d}$$Solving for d, we get:$$d=\frac{3\lambda L}{d}$$$$d=\frac{3(600.0	imes10^{-9}	ext{ m})(0.50	ext{ m})}{0.80	imes10^{-3}	ext{ m}}=1.125	imes10^{-3}	ext{ m}=1.125	ext{ mm}$$Therefore, the width of the slit is 1.125 mm, which is in the range of values listed in choice B.