f 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 = 17 µm. The slit spacing is d = 0.2 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
f 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 = 17 µm. The slit spacing is d = 0.2 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, so m = 1. Substituting the given values:
θ = 1 * 650 * 10^-9 m / 17 * 10^-6 m = 0.0382 radians
This is the angular position of the first minimum in the diffraction pattern, which corresponds to the point where the intensity of the bright fringes drops to zero.
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 need to find the value of n for which θ is just less than 0.0382 radians. Rearranging the formula and substituting the given values:
n = θ * d / λ = 0.0382 radians * 0.2 * 10^-3 m / 650 * 10^-9 m = 11.75
Since n must be an integer, we round down to the nearest whole number, which is 11. Therefore, 11 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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