Coherent light with a wavelength of λ = 500 nm is sent through two parallel slits, each having a width, a = 0.700 µm. The distance between the centres of the slits, d = 2.80 µm. The screen has a semi-cylindrical shape, with its axis at the midline between the slits. Determine the number of maxima.a.12.b.11.c.13.d.10.
Question
Coherent light with a wavelength of λ = 500 nm is sent through two parallel slits, each having a width, a = 0.700 µm. The distance between the centres of the slits, d = 2.80 µm. The screen has a semi-cylindrical shape, with its axis at the midline between the slits. Determine the number of maxima.a.12.b.11.c.13.d.10.
Solution
To solve this problem, we need to use the formula for the angular position of the maxima in a double-slit interference pattern:
mλ = d sin θ
where m is the order of the maximum, λ is the wavelength of the light, d is the distance between the slits, and θ is the angle between the central maximum (m = 0) and the mth maximum.
We want to find the maximum possible value of m, which occurs when sin θ = 1. So we have:
m_max λ = d
Solving for m_max gives:
m_max = d / λ
Substituting the given values (converting them to the same units):
m_max = (2.80 x 10^-6 m) / (500 x 10^-9 m) = 5.6
Since m must be an integer, the maximum possible value of m is 5. However, this is for one side of the central maximum. The total number of maxima, including the central maximum, is therefore 2m + 1 = 2(5) + 1 = 11.
So, the correct answer is (b) 11.
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