Single Slit Diffraction is a fundamental concept in wave optics that explains how light behaves as a wave when passing through a narrow slit. When coherent light (like a laser) goes through a single narrow slit, the waves spread out, and their interaction creates a pattern on a screen placed some distance away. This phenomenon, known as diffraction, leads to the formation of alternating bright and dark regions, showcasing the wave nature of light.
In this article, we'll learn core concepts, types, and practical applications of Single Slit Diffraction, aiming to simplify and explore its patterns and formulas.
Diffraction is defined as the phenomenon in which light bends around the corners of an obstacle whose size is comparable to the wavelength of the light
What is Single Slit Diffraction?
Single Slit Diffraction occurs when light waves pass through a narrow aperture or slit. As the waves spread out, they interfere with each other, resulting in an interference pattern on a screen placed some distance away. This pattern includes a central maximum, secondary maxima, and minima, showing the bending and spreading of light waves around the edges of the slit.
Central Maximum
The central maximum is the brightest region in a single slit diffraction pattern, appearing at the center of the screen. It's formed due to constructive interference, where light waves traveling straight through the slit reinforce each other, resulting in a more intense central peak.
The formula for Central Maximum is:
sin θ = nλ/a
where,
- θ is the Angle from the Center to Maxima
- n is the Order of the Maxima (for central maximum, it is 1)
- λ is the Wavelength of the incident light
- a is the Width of the slit
Path Difference
Path difference describes the variation in distances traveled by light waves from different points on the slit to a specific spot on the screen. It plays a critical role in determining how light waves interact—whether constructively or destructively—at that point. Understanding Single Slit Diffraction offers insights into the intriguing behavior of light waves as they pass through small openings, producing distinct patterns that reveal their wave-like nature.
The formula for path difference in single slit diffraction is:
L = d⋅sin(θ)
where,
- d represents the width of the slit
- θ is the angle of diffraction
Minima Position
The spots where we see darkness or dimness between the bright lines in a single slit diffraction pattern are called minima. They happen because some light waves meet at certain places on the screen and mix in a way that makes those spots look darker. This darkness occurs because these mixed light waves cancel each other out, resulting in reduced brightness or darkness in those particular areas.
The formula for Minima is:
a.sin θ= nλ
where,
- θ is the Angle to Minima
- n is the Order of Minima
- λ is the Wavelength of the Incident light
- a is the Width of the Slit
Position of Secondary Maxima
Secondary maxima are smaller and less intense bright areas close to the brightest spot in a single slit diffraction pattern. They appear because some diffracted waves change direction around the edges of the slit, creating slightly brighter regions away from the brightest spot. This happens because these waves add up and reinforce each other at specific angles, making these areas a bit brighter than their surroundings.
Intensity Distribution Curve (Pattern)
In Diffraction by single slit intensity distribution curve shows how the brightness changes across the diffraction pattern. It explains how bright different angles or spots on the screen are. The curve usually has a highest point in the middle (which represents the central maximum) and then the brightness decreases gradually towards the secondary maxima and minima. This means that the brightest point in the middle is usually much brighter than the slightly less bright areas around it.
The diffraction single by single slit can be be best understood by the mathematical formula also called single slit diffraction formula.
The intensity distribution of diffracted light can be expressed by the single slit diffraction formula:
I(θ) = I0 (sin(β)/ β)2
where,
- I(θ) is the intensity at an angle θ
- I0 is the intensity at the center of the diffraction pattern
- β is the phase difference between waves arriving at different points on the screen
Single Slit Diffraction vs Double Slit Diffraction
Diffraction of light can occur through single slit and double slit. However, the pattern observed in different in both the cases. The difference between single slit diffraction and double slit diffraction is tabulated below:
Single Slit Diffraction | Double Slit Diffraction |
|---|
Involves a single narrow slit. | Involves two adjacent slits. |
Results in an interference pattern with a central maximum, secondary maxima, and minima. | Displays multiple interference patterns with a series of bright and dark fringes. |
Pattern created due to the diffraction of light waves passing through a single slit. | Pattern results from the interference of light waves from both slits. |
Shows a central maximum and secondary maxima and minima. | Shows multiple interference patterns that incorporate characteristics of both single-slit and double-slit interference. |
Exhibits a central maximum and decreasing intensity on either side with secondary maxima. | Shows alternating bright and dark fringes, with the central maximum being more pronounced and subsequent maxima decreasing in intensity. |
Also, Check
Single Slit Diffraction Solved Examples
Example 1: A single-slit diffraction experiment uses light of wavelength λ = 600 nm and a slit width of a = 0.1 mm. Calculate the angular position (in degrees) of the first minimum.
Solution:
a = 0.1 mm = 0.1 × 10(-3) m
λ = 600 nm = 600 × 10(-9) m
For the first minimum, n = 1
Formula for the angular position of the first minimum: sin(θ) = n λ / a
Substituting values: sin(θ) = (1) × (600 × 10(-9)) / (0.1 × 10(-3))
Calculating: sin(θ) ≈ 0.006
Hence, θ ≈ 0.6 degrees
Example 2: In a single-slit diffraction experiment, light of wavelength λ = 500 nm produces the first minimum at an angle x = 0.04 cm from the central maximum. Determine the slit width (in mm).
Solution:
Given:
λ = 500 nm = 500 × 10-9m
x = 0.04 cm = 0.04 × 10-2 m
For the first minimum, n = 1
Formula for slit width: a = n × λ × x / sin(θ)
sin(θ) = x / D (D is the distance between the slit and the screen)
Substituting : a = (1) × (500 × 10(-9)) × (0.04 × 10-2 )/ (sin(θ))
Sin(θ) ≈ θ (for small angles), therefore a ≈ 5 × 10-3 m or 0.05 mm
Example 3. Light of wavelength λ = 500 nm passes through a slit of width a = 0.1 mm, creating a diffraction pattern on a screen. If the first minimum is observed at a distance x = 5 mm from the central maximum, what is the distance (in mm) between the screen and the slit?
Solution:
Given:
λ = 500 nm = 500 × 10-9m
a = 0.1 mm = 0.1 × 10-3m
x = 5 mm = 5 × 10-3m
For the first minimum, n = 1
Formula for distance: D = n λ x / a
Substituting values: D = (1) × (500 × 10-9) × (5 × 10-3) / (0.1 × 10-3)
D ≈ 6.4 mm
Example 4. For a single-slit diffraction pattern, light of wavelength λ = 500 nm produces the first minimum at an angular separation of Δθ = 60 degrees. Calculate the slit width (in mm).
Solution:
λ = 500 nm = 500 × 10-9m
Δθ = 60 degrees
For the central maximum, n = 1
Formula for slit width: a = n × λ / sin(Δθ)
Substituting values: a = (1) × (500 × 10-9) / sin(60 degrees)
Calculating: a ≈ 3.33 × 10-3m or 0.033 mm
Example 5: A monochromatic light of wavelength λ = 600 nm passes through a single slit and produces a diffraction pattern on a screen. If the angular width of the central maximum is 10 degrees, determine the width of the slit.
Solution:
Given:
Wavelength (λ) = 600 nm = 600 × 10-9 m
Angular width of the central maximum = 10 degrees
For the central maximum, n = 1
Formula for the width of the slit: a = n × λ / sin(Δθ)
Substituting values:
a = (1) × (600 × 10-9) / sin(10 degrees)
Calculating:
a ≈ 3.47 × 10-5 m or 34.7 μm
Practice Problems on Single Slit Diffraction
Q1: A single-slit diffraction pattern is formed on a screen 2 meters away. If a light of wavelength 500 nm produces the first minimum at an angle of 30 degrees, what is the width of the slit?
Q2: Light with a wavelength of 600 nm passes through a single slit and produces a diffraction pattern. If the angular width of the central maximum is 20 degrees, calculate the width of the slit.
Q3: In a single-slit diffraction experiment, if the second minimum is observed at an angle of 45 degrees and the wavelength of light used is 450 nm, find the width of the slit.
Q4: A diffraction pattern is observed on a screen placed 1.5 meters away from a single slit. If light of wavelength 700 nm produces the first minimum at an angle of 45 degrees, determine the width of the slit.
Q5: When monochromatic light with a wavelength of 400 nm passes through a single slit, the first minimum is observed at an angle of 15 degrees. What is the width of the slit?
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