In a single-slit diffraction pattern, what is the angular width of the central maximum if the slit width is 0.5 mm and light of wavelength 500 nm is used?
Practice Questions
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Q1
In a single-slit diffraction pattern, what is the angular width of the central maximum if the slit width is 0.5 mm and light of wavelength 500 nm is used?
0.1 rad
0.2 rad
0.3 rad
0.4 rad
Angular width = 2λ/a = 2(500 nm)/(0.5 mm) = 2(500 x 10^-9)/(0.5 x 10^-3) = 0.002 rad.
Questions & Step-by-step Solutions
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Q
Q: In a single-slit diffraction pattern, what is the angular width of the central maximum if the slit width is 0.5 mm and light of wavelength 500 nm is used?
Solution: Angular width = 2λ/a = 2(500 nm)/(0.5 mm) = 2(500 x 10^-9)/(0.5 x 10^-3) = 0.002 rad.
Steps: 6
Step 1: Identify the given values. The slit width (a) is 0.5 mm and the wavelength (λ) is 500 nm.
Step 2: Convert the units to be consistent. Convert 0.5 mm to meters: 0.5 mm = 0.5 x 10^-3 m.
Step 3: Convert 500 nm to meters: 500 nm = 500 x 10^-9 m.
Step 4: Use the formula for angular width of the central maximum in a single-slit diffraction pattern: Angular width = 2λ/a.
Step 5: Substitute the values into the formula: Angular width = 2(500 x 10^-9 m) / (0.5 x 10^-3 m).
Step 6: Calculate the result: Angular width = 2(500 x 10^-9) / (0.5 x 10^-3) = 0.002 rad.
