PHY2606
Assignment 4 Solutions
Chapter 12
Unique Number: 685128
Due Date: 21st July 2025
, Assignment 4: Waves in Optical Systems
Due: 21st July 2025
Question 1: Double-Slit Interference and Diffraction
Problem Statement
(a) Monochromatic light illuminates two slits separated by 1.2 mm, creating an interfer-
ence pattern on a screen 3.6 m from the slits. The distance between the third and sixth
dark fringes on the screen is 5.3 mm. Calculate the wavelength of the light.
(b) The width of each slit in part (a) is a = 0.15 mm. Determine the width of the central
diffraction maximum on the screen and how many bright fringes (interference maxima)
are contained within it.
Part (a): Wavelength of the Light
Step 1: Understand the Interference Pattern and Dark Fringes
In a double-slit experiment, dark fringes occur at:
1
d sin θ = m + λ
2
where d = 1.2 mm = 1.2 × 10−3 m, λ is the wavelength, m is the order, and θ is the angle.
For small angles, sin θ ≈ Ly , where y is the fringe position and L = 3.6 m. The position
of the m-th dark fringe is:
m + 21 λL
ym =
d
The distance between the third (m = 2) and sixth (m = 5) dark fringes is ∆y = 5.3 mm =
5.3 × 10−3 m.
Step 2: Calculate Positions of Dark Fringes
For the third dark fringe (m = 2):
2.5λL
y2 =
d
For the sixth dark fringe (m = 5):
5.5λL
y5 =
d
1
Assignment 4 Solutions
Chapter 12
Unique Number: 685128
Due Date: 21st July 2025
, Assignment 4: Waves in Optical Systems
Due: 21st July 2025
Question 1: Double-Slit Interference and Diffraction
Problem Statement
(a) Monochromatic light illuminates two slits separated by 1.2 mm, creating an interfer-
ence pattern on a screen 3.6 m from the slits. The distance between the third and sixth
dark fringes on the screen is 5.3 mm. Calculate the wavelength of the light.
(b) The width of each slit in part (a) is a = 0.15 mm. Determine the width of the central
diffraction maximum on the screen and how many bright fringes (interference maxima)
are contained within it.
Part (a): Wavelength of the Light
Step 1: Understand the Interference Pattern and Dark Fringes
In a double-slit experiment, dark fringes occur at:
1
d sin θ = m + λ
2
where d = 1.2 mm = 1.2 × 10−3 m, λ is the wavelength, m is the order, and θ is the angle.
For small angles, sin θ ≈ Ly , where y is the fringe position and L = 3.6 m. The position
of the m-th dark fringe is:
m + 21 λL
ym =
d
The distance between the third (m = 2) and sixth (m = 5) dark fringes is ∆y = 5.3 mm =
5.3 × 10−3 m.
Step 2: Calculate Positions of Dark Fringes
For the third dark fringe (m = 2):
2.5λL
y2 =
d
For the sixth dark fringe (m = 5):
5.5λL
y5 =
d
1
