In an experiment using sodium light (λ=589 nm), the slits are 1.2 μm wide. You notice that every fourth fringes missing.
a) What is the slit separation?
b) How many fringes appear on the screen?
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To answer these questions, we can use the concept of interference in a double-slit experiment. The interference pattern is created by the constructive and destructive interference of light waves passing through the two slits.
a) To find the slit separation, we need to determine the wavelength of the light, the width of the slits, and the missing fringe pattern.
Given:
- Wavelength (λ) = 589 nm = 589 × 10^(-9) meters
- Width of the slits (a) = 1.2 μm = 1.2 × 10^(-6) meters
- Missing fringes (n) = every fourth fringe
In an interference pattern, the condition for constructive interference is given by:
d sinθ = mλ
Where:
- d is the slit separation (the distance between the centers of the two slits)
- θ is the angle between the central maximum and the mth-order fringe
- m is the order of the fringe
- λ is the wavelength of light
Since we know that every fourth fringe is missing, it means that we can find the missing m values by counting the fringes (starting from the central maximum). The missing fringes will correspond to values of m that are multiples of 4.
Starting from m = 0 (the central maximum), we find that m = 0, 1, 2, 3 are the first four fringes. Therefore, the m values corresponding to the missing fringes are m = 4, 8, 12, 16, etc.
Now let's consider the m = 4 fringe:
d sinθ = mλ
d sinθ = 4λ
Since we want to find the slit separation (d), we rearrange the equation:
d = (4λ) / sinθ
We know the wavelength (λ = 589 × 10^(-9) meters), so the next step is to find the angle θ. This can be done by considering the distance x between the central maximum and any of the observed fringes.
The distance between the central maximum and the mth fringe can be given by:
x = L tanθ
Where:
- L is the distance between the slits and the screen
- θ is the angle between the central maximum and the mth-order fringe
Since we don't have the values of x and L, we cannot directly find θ. However, we can make an approximation by considering the small-angle approximation, which is valid for small angles such as those observed in the interference pattern.
Using the small-angle approximation, we have:
tanθ ≈ sinθ ≈ θ (in radians)
Therefore, we can approximate the angle as:
θ ≈ x / L
Since we are given the missing fringes (every fourth), let's assume that the missing fringe occurs at m = 4. So, we can consider x as the distance between the central maximum and the missing fourth fringe.
Now, we apply the small-angle approximation:
θ ≈ x / L
Substituting this approximation into the equation for the slit separation:
d = (4λ) / sinθ ≈ (4λ) / (x / L) = (4Lλ) / x
We are given the width of the slits (a) = 1.2 × 10^(-6) meters, which is very small compared to the distance x. Therefore, we can consider that the distance x is approximately equal to the width of the central maximum.
For a double-slit interference pattern, the width of the central maximum can be approximated as:
w ≈ 2λL / a
Substituting this approximation into the equation for the slit separation:
d ≈ (4Lλ) / x ≈ (4Lλ) / w ≈ (4Lλ) / (2λL / a) = (2aL) / L = 2a
Therefore, the slit separation is approximately equal to 2 times the width of the slits:
d ≈ 2a = 2 × 1.2 × 10^(-6) meters
b) To find the number of fringes appearing on the screen, we need to consider the interference pattern and take into account the missing fringes.
Since every fourth fringe is missing, the number of observed fringes (N) can be calculated using the formula:
N = (n + 1) × 4
Where:
- n is the number of missing fringes. In this case, n = 0 since there are no missing fringes before the first observed fringe.
- 4 is the number of fringes between every missing fringe.
Therefore, the number of fringes appearing on the screen is:
N = (0 + 1) × 4 = 4 fringes
In summary:
a) The slit separation is approximately 2 × 1.2 × 10^(-6) meters.
b) The number of fringes appearing on the screen is 4 fringes.