If light illuminates a double slit an interference pattern of alternating bright and dark spots is seen. The intensity of the bright spots is brighter at the centre due to the width of the slits. The bright interference fringes will be readily observed inside the broad central diffraction maximum. Outside this they will be much fainter.A laser of wavelength λ = 650 nm is shone on a pair of slits each of width a = 14 µm. The slit spacing is d = 0.3 mm. The resulting interference pattern is projected on to a screen a distance L= 2.6 m from the double slit.Counting out from the (n= 0) central bright interference fringe, how many more bright interference fringes will be seen before the intensity drops to zero at the edge of the central diffraction maximum.Express your answer to the nearest whole number.
Question
If light illuminates a double slit an interference pattern of alternating bright and dark spots is seen. The intensity of the bright spots is brighter at the centre due to the width of the slits. The bright interference fringes will be readily observed inside the broad central diffraction maximum. Outside this they will be much fainter.A laser of wavelength λ = 650 nm is shone on a pair of slits each of width a = 14 µm. The slit spacing is d = 0.3 mm. The resulting interference pattern is projected on to a screen a distance L= 2.6 m from the double slit.Counting out from the (n= 0) central bright interference fringe, how many more bright interference fringes will be seen before the intensity drops to zero at the edge of the central diffraction maximum.Express your answer to the nearest whole number.
Solution
To solve this problem, we need to use the formula for the angular position of dark fringes in single-slit diffraction, which is given by:
sin(θ) = m * λ / a
where:
- m is the order of the fringe,
- λ is the wavelength of the light, and
- a is the width of the slit.
We are looking for the first dark fringe (m = 1) that will mark the boundary of the central maximum.
First, we need to convert all the given measurements to the same units. Here, we will convert everything to meters:
λ = 650 nm = 650 * 10^-9 m a = 14 µm = 14 * 10^-6 m
Substituting these values into the formula, we get:
sin(θ) = 1 * (650 * 10^-9 m) / (14 * 10^-6 m) = 0.0464
Now, we need to find the position of this dark fringe on the screen. We can use the small angle approximation (tan(θ) ≈ sin(θ) when θ is small), which gives us:
y = L * tan(θ) ≈ L * sin(θ)
where:
- y is the position on the screen, and
- L is the distance from the slit to the screen.
Substituting the given values, we get:
y = 2.6 m * 0.0464 = 0.12064 m
Now, we need to find the number of bright fringes within this distance. The position of bright fringes in double-slit interference is given by:
y = m * λ * L / d
where:
- m is the order of the fringe,
- λ is the wavelength of the light,
- L is the distance from the slit to the screen, and
- d is the spacing between the slits.
Rearranging this formula to solve for m, we get:
m = y * d / (λ * L)
Substituting the given values, we get:
m = 0.12064 m * (0.3 * 10^-3 m) / ((650 * 10^-9 m) * 2.6 m) = 21.2
Rounding to the nearest whole number, we get 21. Therefore, 21 more bright interference fringes will be seen before the intensity drops to zero at the edge of the central diffraction maximum.
Similar Questions
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