A pair of narrow, parallel slits separated by 0.234 mm are illuminated by the green component from a mercury vapor lamp (ë=531.5 nm). The interference pattern is observed on a screen 1.21 m from the plane of the parallel slits.
a)Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
b)Calculate the distance between the first and second dark bands in the interference pattern.
To answer these questions, we can use the principles of interference and diffraction.
a) The distance from the central maximum to the first bright region on either side of the central maximum can be calculated using the formula for the position of bright fringes in a double-slit interference pattern:
y = λL / d
Where:
- y is the distance from the central maximum to the bright region
- λ is the wavelength of light (531.5 nm or 5.315 x 10^-7 m in this case)
- L is the distance from the slits to the screen (1.21 m in this case)
- d is the distance between the two slits (0.234 mm or 2.34 x 10^-4 m in this case)
Plugging in the values into the formula, we get:
y = (5.315 x 10^-7 m) x (1.21 m) / (2.34 x 10^-4 m)
y = 2.733 x 10^-3 m
So, the distance from the central maximum to the first bright region on either side is approximately 2.733 mm.
b) The distance between the first and second dark bands in the interference pattern can be calculated using the same formula, but for dark fringes:
y = (m + 1/2) λL / d
Where:
- y is the distance between the dark bands
- m is the difference between the order of the dark fringe and the central maximum (m = 1 in this case)
- λ is the wavelength of light (again, 5.315 x 10^-7 m)
- L is the distance from the slits to the screen (1.21 m)
- d is the distance between the two slits (0.234 mm)
Plugging in the values into the formula, we get:
y = [(1 + 1/2) x (5.315 x 10^-7 m) x (1.21 m)] / (2.34 x 10^-4 m)
y = 1.367 x 10^-3 m
So, the distance between the first and second dark bands is approximately 1.367 mm.