A pair of narrow parallel slits separated by a
distance of 0.253 mm are illuminated by the
green component from a mercury vapor lamp
(λ = 545.1 nm).
What is the angle from the central maximum to the first bright fringe on either side of
the central maximum?
Answer in units of
◦
tanα = x(max1)/L = λ/d =545.1•10^-9/0.253•10^-3 =2.15•10^-3
α = 0.123 o
Im sorry to ask again but what equation did you use? Can you list the steps you did please?
So I can answer the next few questions by myself.
To find the angle from the central maximum to the first bright fringe, we can use the concept of diffraction.
First, let's define the variables:
d = distance between the slits (0.253 mm)
λ = wavelength of light (545.1 nm)
The angle from the central maximum to the first bright fringe can be found using the formula:
sin(θ) = m * λ / d
Where:
θ = angle
m = order number of the fringe (for the first bright fringe, m = 1)
Substituting the values into the formula:
sin(θ) = (1 * 545.1 nm) / (0.253 mm)
Now, we need to convert the units to be consistent. We know that 1 mm = 10^6 nm, so:
sin(θ) = (1 * 545.1 nm) / (0.253 mm * 10^6 nm/mm)
Next, we calculate the value of sin(θ):
sin(θ) = (1 * 545.1 nm) / (0.253 * 10^6 nm)
sin(θ) = 2.154 * 10^-4
Now, we can find the value of θ by taking the inverse sin of sin(θ):
θ = arcsin(2.154 * 10^-4)
Using a scientific calculator, we find:
θ ≈ 0.0124 radians
To convert this angle to degrees, we multiply by 180/π:
θ ≈ 0.0124 * (180/π) degrees
θ ≈ 0.712 degrees
Therefore, the angle from the central maximum to the first bright fringe on either side of the central maximum is approximately 0.712 degrees.