The Diffraction Of Light Due To A Single Slit

What is Diffraction?

Diffraction is defined as the bending of light around corners such that it spreads out and illuminates areas where a shadow is expected. In general, it is hard to separate diffraction from interference since both occur simultaneously. When the double slit in Young’s experiment is replaced by a single narrow slit, a broad pattern with a bright region at the centre is seen. On both sides of the centre, there are alternating dark and bright regions. The intensity becomes weaker away from the centre. In this article, we discuss the single slit diffraction of light in detailed manner.

Single Slit Diffraction

In the single slit experiment, we can observe the bending phenomenon of light or diffraction that causes light from a coherent source interfere with itself and produce a distinctive pattern on the screen called the diffraction pattern. Diffraction is evident when the sources are small enough that they are relatively the size of the wavelength of light. You can see this effect in the diagram below. For large slits, the spreading out is small and generally unnoticeable.

Single Slit
For we shall assume the slit width a << D. x`D is the separation between slit and source.

Single Slit
We shall identify the angular position of any point on the screen by ϑ measured from the slit centre which divides the slit by \(\frac{a}{2}\) lengths. To describe the pattern, we shall first see the condition for dark fringes. Also, let us divide the slit into zones of equal widths \(\frac{a}{2}\). Let us consider a pair of rays that emanate from distances  \(\frac{a}{2}\) from each other as shown.

Single Slit
The path difference exhibited by the top two rays shown is:

\(\Delta L=\frac{a}{2}\sin \Theta\)

Remember that this is a calculation valid only if D is very large. For more details about the approximation check out our article on the Young’s Double Slit experiment.

We can consider any number of ray pairings that start from a distance \(\frac{a}{2}\) from one another such as the bottom two rays in the diagram. Any arbitrary pair of rays at a distance \(\frac{a}{2}\)can be considered. We shall see the importance of this trick in a moment.

For a dark fringe, the path difference must cause destructive interference; the path difference must be out of phase by \(\frac{\lambda}{2}\). (λ is the wavelength)

For the first fringe,

ΔL = \(\frac{\lambda}{2}\) = \(\frac{a}{2}\sin \Theta\)

λ = a sin θ

For a ray emanating from any point in the slit, there exists another ray at a distance \(\frac{a}{2}\) that can cause destructive interference.

Thus, at θ = sin−1λa, there is destructive interference as any ray emanating from a point has a counterpart that causes destructive interference. Hence, a dark fringe is obtained.

For the next fringe, we can divide the slit into 4 equal parts of a/4 and apply the same logic. Thus, for the second minima

\(\frac{\lambda}{2}=\frac{a}{4}\sin\Theta\) \(2\lambda =a\sin \Theta\)

Similarly, for the nth fringe, we can divide the slit into 2n parts and use this condition as

nλ = a sin θ

The central maximum

The maxima lie between the minima and the width of the central maximum is simply the distance between the 1st order minima from the centre of the screen on both sides of the centre.

The position of the minima given by y (measured from the centre of the screen) is:

tanθ≈θ≈yD

For small ϑ, sin θ≈θ ,

λ = asin θ≈aθ

θ = yD = λa

y = λDa

The width of the central maximum is simply twice this value

Width of central maximum = 2λDa

Angular width of central maximum = 2θ = 2λa

The diffraction pattern and intensity graph are shown below.

Diffraction Pattern
Stay tuned with BYJU’S and learn other important concepts like thermodynamics and doppler effect with the help of interactive video lessons.

Single Slit Diffraction Important Questions

Q1. How are a phase difference (Φ) and path difference (δ) related?

Ans: Following is the equation explaining the relationship between phase difference (Φ) and path difference (δ):

\(\Phi =\frac{2\pi }{\lambda }\delta\)

Q2. What is phase difference?

Ans: Phase difference is defined as the difference between any two waves or the particles having the same frequency and starting from the same point. It is expressed in degrees or radians.

Q3. What is the condition for constructive interference?

Ans: The condition for constructive interference is that the path difference should be equal to an integral multiple of the wavelength.

Q4. What is the condition for destructive interference?

Ans: The condition for destructive interference is that the path difference should be equal to odd integral multiple of half wavelength.

Q5. What is temporal coherence?

Ans: Temporal coherence is known as the correlation between the field at a point and the field at the same point at the later time.

Q6. What is Newton’s rings experiment equation?

Ans: The equation of Newton’s rings experiment is given as:

\(\lambda =\frac{D_{n}^{2}-D_{m}^{2}}{4R(n-m)}\)

Q7. Which lamp is used as a source of light to demonstrate Newton’s rings experiment?

Ans: Sodium vapour lamp is used.

Q8. What is diffraction?

Ans: Diffraction is the phenomenon of bending of light around an obstacle.

Q9. What is Compton effect?

Ans: Compton effect is defined as the increase in the wavelength of the X-rays and the gamma rays which occurs when they are scattered.

Q10. How does rainbow form?

Ans: the formation of a rainbow is due to refraction and reflection of sunlight by the water droplets.

Q11. On what principle does optics fibres work on?

Ans: Optics fibres work on the principle of total internal reflection.

Q12. What is total internal reflection?

Ans: Total internal reflection is defined as the complete reflection of light rays within the medium.

Q13. What is the relationship between fringe width and the angle of the wedge?

Ans: The relationship between fringe width and the angle of the wedge is given as:

\(\beta =\frac{\lambda }{2\theta }\)

Q14. What is fringe width?

Ans: Fringe width is defined as the distance between any two consecutive bright or dark fringes.

Q15. What is the unit of wavelength?

Ans: Metre (m)


Practise This Question

The two slits in Young's interference experiment have widths in the ratio n:1. The ratio of the intensities of the maxima and minima in the interference pattern is