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 = 27 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 = 27 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 the minima in the diffraction pattern of a single slit, which is given by:
θ = m * λ / a
where:
- m is the order of the minima,
- λ is the wavelength of the light, and
- a is the width of the slit.
We are looking for the first minimum (m = 1), so we can substitute these values into the formula to find the angle:
θ = 1 * 650e-9 m / 27e-6 m = 0.02407 radians
Next, we use the formula for the angular position of the maxima in the interference pattern of a double slit, which is given by:
θ = n * λ / d
where:
- n is the order of the maxima, and
- d is the spacing between the slits.
We can rearrange this formula to solve for n:
n = θ * d / λ
Substituting the values we have:
n = 0.02407 radians * 0.3e-3 m / 650e-9 m = 11.1
Since we can't have a fraction of a fringe, we round this to the nearest whole number. Therefore, 11 more bright interference fringes will be seen before the intensity drops to zero at the edge of the central diffraction maximum.
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