wavelength"\\lambda = 725\\, \\text{nm}"
Distance from slit to screen "L = 1.25\\, \\text{m}"
Width of the slit "d = 18.5 \\, \\mu\\text{m}"
Find the angle for the first minimum:
"d \\sin \\theta = \\lambda"
It is small, therefore:
"\\sin \\theta = \\frac{\\lambda}{d} \\approx \\frac{x}{L} \\Rightarrow x = \\frac{\\lambda L}{d}"
The width of the central maximum :
b)
"\\Delta = 2x = \\frac{2\\lambda L}{d} = 9.8\\, \\text{cm}"
a)
"2\\theta = \\frac{2\\lambda}{d}\\, \\text{rad} =\\frac{2\\lambda}{d} \\cdot 180\/\\pi = 4.5^\\circ"
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