A double slit diffraction pattern is formed on a distant screen using mercury green light of
wavelength 546.1 nm. Each slit has a width of 0.1 mm. The pattern shows that the 4th order
interference maximum is missing.
a) What is the separation of the slits?
b) What is the intensity of the 2nd order interference maximum relative to the zero order
maximum?
c) Show that the number of bright fringes seen under the central diffraction peak is always
given by 2(a/b)-1, where a is the separation of the slits.
a) To find the separation of the slits, we can use the formula for the angle of the interference maximum:
sin(θ) = mλ/d
where θ is the angle of the interference maximum, m is the order of the maximum, λ is the wavelength of light, and d is the separation of the slits.
Since the 4th order maximum is missing, we can use the angle of the 3rd order maximum to calculate the separation of the slits.
For the 3rd order maximum, m = 3 and λ = 546.1 nm = 0.5461 μm.
Using the given formula, we have:
sin(θ) = 3(0.5461 μm) / d
Now, we need to convert the slit width into meters to be consistent with the units of the wavelength:
0.1 mm = 0.1 × 10^{-3} m.
Substituting the values into the equation, we have:
sin(θ) = 3(0.5461 × 10^{-6} m) / (0.1 × 10^{-3} m)
Now, solve for d:
d = 3(0.5461 × 10^{-6} m) / (0.1 × 10^{-3} m)
d ≈ 16.3833 μm
Therefore, the separation of the slits is approximately 16.3833 μm.
b) To find the intensity of the 2nd order interference maximum relative to the zero order maximum, we can use the formula:
I_m = I_0 * (sin(θ)/θ)^2 * (sin(Nθ)/sin(θ))^2
where I_m is the intensity at the mth order maximum, I_0 is the intensity at the zero order maximum, θ is the angle of the interference maximum, and N is the number of slits.
Since we are looking for the 2nd order maximum, m = 2.
Using the given values, we have λ = 546.1 nm = 0.5461 μm, and d = 16.3833 μm.
To calculate the intensity ratio, we need to find the angle θ for both the zero and 2nd order maxima.
For the zero order maximum, m = 0, so sin(θ_0) = 0 and θ_0 = 0 degrees.
For the 2nd order maximum, we have m = 2:
sin(θ_2) = (2λ) / d
θ_2 = arcsin((2 × 0.5461 × 10^{-6} m) / 16.3833 × 10^{-6} m)
Now, we can substitute the values into the intensity formula:
I_2 = I_0 * (sin(θ_2)/θ_2)^2 * (sin(Nθ_2)/sin(θ_2))^2
Substituting θ_2 = arcsin((2 × 0.5461 × 10^{-6} m) / 16.3833 × 10^{-6} m) into the formula, we can calculate the intensity ratio.
c) To show that the number of bright fringes seen under the central diffraction peak is given by 2(a/b) - 1, we need to understand the concept of a central diffraction peak.
The central diffraction peak appears when the path length difference between the two slits is zero. This occurs when the angle θ is equal to zero degrees.
For the central peak, m = 0. Therefore, sin(θ) = 0.
Using the formula for sin(θ) = mλ/d, we can rewrite it as:
mλ = d * sin(θ)
Since sin(θ) = 0 for the central peak, we have:
mλ = 0
This means that the only bright fringe seen under the central diffraction peak is m = 0. Thus, the number of bright fringes seen under the central diffraction peak is given by:
2(0) - 1 = -1
Therefore, there are no bright fringes under the central diffraction peak.