27.3 TWO-SLIT INTERFERENCE
27.3.1 Young’s Two-slit Experiment
1 Experimental setup
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Figure 27.8 The diagram shows the arrangement used to produce a stationary light interference pattern, including a coherent light source, two slits, and a screen where alternating bright and dark fringes are formed. |
Figure 27.8 shows a simple diagram of the components used in producing a stationary interference of light. The components are as follows:
1 Apparatus
- A monochromatic light source.
- A single slit S0. The width of the slit is 0.1 mm – 0.3 mm.
- Double slits S1 and S2. Separation of slits a is 0.2 – 0.4 mm. Width of slit is 0.1 mm – 0.3 mm. Distance d between S0 and the double slits is 20 cm – 40 cm.
- A screen. Distance of screen D from the double slits is 80 cm – 150 cm.
- A travelling microscope to view the interference fringes formed on the screen.
2 Principle
- Light that passes through each slit, S1 and S2, undergoes diffraction because the slit width is very small.
- The double slits behave like a pair of coherent light sources. The two beams originate from the same single light source. The waves from the two beams have the same amplitude, same frequency and a constant phase difference.
- The two diffracted light beams travel through a distance large enough for them to have a chance to overlap on the screen.
- At each place on the screen where a bright fringe is formed, constructive interference takes place. At a place where a dark fringe is formed, destructive interference takes place.
- An interference pattern consisting of a set of parallel equally spaced bright fringes separated by dark fringes is formed on the screen.
3 Formula for Young’s interference pattern
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| Figure 27.9 The diagram shows point P as the m-th bright fringe from the central fringe O. The path difference at P determines the interference condition, where constructive interference occurs when p = mλ. |
Figure 27.9 shows the setup for Young’s two-slit experiment. P is the m-th bright fringe from the central bright fringe O. The path difference at P is
PS2 − PS1 = mλ …… (1)
where λ is the wavelength of the light used. This fringe is at distance xm from O.
Refer to ΔOPQ. We have
tan θ = xm / D
The angle θ is very small because xm is very short. Therefore, tan θ ≈ θ where θ is in radian.
θ ≈ xm / D …… (2)
Refer to ΔS1S2R where S1R is perpendicular to PS2.
sin θ = RS2 / S1S2
Since θ is very small, sin θ ≈ θ.
θ ≈ RS2 / a …… (3)
Equating (2) and (3),
RS2 / a = xm / D
RS2 = (a xm) / D …… (4)
But RS2 is the path difference at point P. From (1), we have
RS2 = path difference for P = mλ
Using (4),
(a xm) / D = mλ
xm = m (λD / a)
Next, consider the (m + 1)-th bright fringe, which is adjacent to fringe P. The distance of this fringe from O is xm+1 where
xm+1 = (m + 1) (λD / a)
The fringe separation y is given by
y = xm+1 − xm
= (m + 1)(λD / a) − m(λD / a)
y = λD / a
EXAMPLE 27.2
In a Young’s double slit experiment, an interference pattern is formed on a screen 1.00 m away from the double slits. The double slits are separated by 0.25 mm and the wavelength of light used is 550 nm.
(a) Why is the central fringe a bright fringe?
(b) Determine the distance from the central bright fringe the distance of
(i) the 5th bright fringe
(ii) the 3rd dark fringe
Answer
(a) The central fringe is equidistant from both slits. The path difference for this fringe is zero. A zero path difference favours constructive interference. Hence, a bright fringe is formed.
y = λD / a
= (550 × 10−9)(1.00) / (0.25 × 10−3) = 2.2 × 10−3 m
2.2 mm
Distance of 5th bright fringe, x5
= 5y
= 5(2.2) = 11 mm
(ii) Distance of 3rd dark fringe = 3 1/2 y
= (3.5)(2.2) = 7.7 mm
EXAMPLE 27.3
Light of wavelength 600 nm is incident upon a Young’s double slits which are 1.0 m from a screen. The slit separation is 0.30 mm. Determine the separation between the 2nd bright fringe and the 4th dark fringe.
Answer
y = λD / a
= (600 × 10−9)(1.0) / (0.30 × 10−3) = 2.0 × 10−3 m = 2.0 mm
The distance d between 2nd bright fringe and 4th dark fringe is
d = 2 1/2 y
= (2.5)(2.0) = 5.0 mm
EXAMPLE 27.4
A beam of light composing of two wavelengths, λ1 and λ2, is used in a Young’s double slit experiment. It is observed that the 4th dark fringe produced by λ1 coincides with the 3rd bright fringe produced by λ2 on a screen. If λ1 has a value of 510 nm, determine the value of λ2.
Answer
4th dark fringe:
y1 = λ1D / a
distance from central fringe x1 = 3 1/2 y1
3rd bright fringe:
y2 = λ2D / a
distance from central fringe x2 = 3y2
Given
x1 = x2
3 1/2 y1 = 3y2
(3/2) λ1D / a = 3 λ2D / a
λ2 = 7/6 λ1
= (7/6)(510) = 595 nm
🔬 Applications of Young’s Double-Slit & Interference
The principles of interference and Young’s experiment are widely used in science, engineering, and modern technology.
🌈 Optical Instruments
Interference is used in devices such as interferometers to measure very small distances, wavelengths, and optical properties with high precision.
🧪 Wavelength Measurement
Young’s double-slit experiment is used to determine the wavelength of light by measuring fringe spacing on a screen.
💻 Fiber Optics & Communication
Interference principles help in signal transmission and minimizing signal loss in fiber optic communication systems.
🎨 Color Formation
The formation of colorful patterns in soap bubbles and thin films is explained using light interference.
🧬 Scientific Research
Used in advanced experiments, including quantum physics, to study the wave nature of light and particles.
📸 Holography
Interference patterns are used to create 3D images in holography technology.
💡 Key Insight
Young’s experiment proves that light behaves as a wave, and interference patterns provide a powerful tool for measurement, imaging, and modern technology.
✨ Conclusion
Young’s Double-Slit Experiment clearly demonstrates the wave nature of light through the formation of interference patterns. By analyzing bright and dark fringes, we can understand how waves interact through constructive and destructive interference.
This experiment not only strengthens the concept of wave behavior but also provides a foundation for many modern technologies such as optical instruments, communication systems, and advanced scientific research.
❓ Frequently Asked Questions
Click each question to reveal the answer
An experiment that demonstrates the wave nature of light by producing an interference pattern of bright and dark fringes.
The central fringe is bright because the path difference is zero, leading to constructive interference.
Bright fringes are caused by constructive interference, while dark fringes are caused by destructive interference.
Fringe spacing is the distance between two consecutive bright or dark fringes, given by y = λD / a.
Coherent sources produce waves with the same frequency, amplitude, and a constant phase difference.
It is used in optical instruments, wavelength measurement, fiber optics, and advanced scientific research.
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