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April 11, 2024We classically think that light always travels in straight lines, but do you know that when light waves pass near an obstacle, they tend to bend around that obstacle and become spread out? When the light comes or passes through an obstacle, **Diffraction of light **takes place.

Diffraction of light is divided mainly into two types: Fraunhofer Diffraction and Fresnel Diffraction. The Diffraction of light is used to transform the light into its spectrum. It is used in holography. This article will provide detailed information on the Diffraction of light. Scroll down to learn more!

As we all know that, light travels in a line. However, when light passes through a tiny hole (dimension comparable to the wavelength of light), there’s a particular amount of spreading of light. When light passes by an obstacle, then it appears to bend around the edges of the obstacle. The phenomenon of bending of light around the corners of small blocks or apertures and its consequent spreading into the regions of the geometrical shadow is termed **Diffraction** of light. Figure (a) and (b) shows the diffraction pattern below:

The bending won’t be noticeable if the opening is much larger than the wavelength of light. So, it is possible when both are equal or closer in size for a considerable amount of bending that can be seen with the naked eye. Diffraction is a general characteristic exhibited by all types of waves, such as sound waves, light waves, water waves, or matter waves.**Essential Condition for observation of Diffraction.**

1. The source of light should be coherent.

2. The wavelength of the light used should be comparable to the size of the obstacle.

Diffraction is mainly of two types:

1. Fraunhofer Diffraction

2. Fresnel Diffraction.**1. Fraunhofer Diffraction:** This occurs when all the rays passing through a narrow slit are approximately parallel to one another. This can be achieved either by placing the light source far from the opening used to create the Diffraction, as shown in the figure below. The important points for Fraunhofer Diffraction are given below:

a. The screen and the source are an infinite distance from each other.

b. The incident wavefronts are horizontal.

c. The plane or horizontal wavefront are connected using a convex lens to produce a diffraction pattern.

d. Plane diffracting waves are conserved using a convex lens to produce a diffracting pattern.

Fig. Fraunhofer Diffraction Pattern

**2. Fresnel Diffraction**:- In Fresnel Diffraction, the source of light and the screen on which the Diffraction pattern is obtained are effectively at a finite distance from the diffracting system. The important points for Fresnel Diffraction are given below:

a. The screen and the source are near to each other.

b. The shapes of Incident wavefronts are spherical or cylindrical.

c. The objects left by the wavefronts are spherical.

d. A convex lens is not needed to converse the spherical wavefronts.

Fig 2. Fresnel Diffraction

When the light coming from a coherent source passes through a finite-size hole having dimensions comparable to the wavelength of light, we observe a diffraction pattern. When the sources are very small to the dimensions of the wavelength, Diffraction is noticeable. A central maximum with some thinner pattern on both sides is formed in this diffraction pattern. The central maximum will have the highest intensity, and the intensity decreases rapidly on either side.

In this topic, we will see how the finite width of slits is the basis for understanding Fraunhofer diffraction.

Fig. Diffraction of Light by a single slit of width *a*

Let us consider waves coming from various portions of the slit in the given figure. Now, According to Huygens’s principle, **each portion of the slit acts as a source of light waves. **So that the light from one portion of the slit can interfere with light from another portion of the slit, and the resultant light intensity will depend on the direction \(θ\) and it can be obtained on a screen.

\(\frac{a}{2}\sin \theta = \frac{\lambda}{2}\)

or when

\(\sin \theta = \frac{\lambda}{a}\)

Again using a similar reason, if we divide the slit into four equal parts, we will notice that the viewing screen is also dark when,

\(\sin \theta = \frac{2 \lambda}{a}\)

In the same way, if we will divide the slit into six equal parts and again, we will notice that darkness occurs on the screen when:

\(\sin \theta = \frac{3 \lambda}{a}\)

Therefore, for destructive interference, we can write the general condition as:

\(\sin \theta = m \frac{\lambda}{a}\)

Where \(m = ±1,\,±2,\,±3\),…

This equation gives the values of \(θ\) for which the diffraction pattern has zero light intensity—that is, when a dark fringe is formed. The intensity distribution of the fringes is shown in Fig 4. We can observe a broad central bright fringe is flanked by much weaker bright fringes, which is alternate with dark fringes. The various dark fringes occur at the values of \(θ\). And each bright fringe peak will lie approximately halfway between its bordering dark fringe minima.

Fig. Intensity distribution for a Fraunhofer Diffraction Pattern

*Q.1. The first diffraction minima due to single slit diffraction are at \(θ = 30°\) for a light of wavelength \(5000\,(Å)\). Then, find the width of the slit?*

** Ans:** The distance of the first diffraction minimum from the central minimum

\(x = \frac{{\lambda D}}{d}\)

Therefore,

\(\sin \theta = \frac{x}{D} = \frac{\lambda }{d}\)

\( \Rightarrow d = \frac{\lambda }{{\sin \theta }}\)

\( \Rightarrow d = \frac{{5000 \times {{10}^{ – 10}}}}{{\sin {{30}^ \circ }}}{\text{m}}\)

\( \Rightarrow d = 1.0 \times 10^{-4}\,\rm{cm}\)

*Q.2. Two spectral lines of sodium \(D_1\) and \(D_2\) have wavelengths of approximately \(5890\,(Å)\) and \(5896\,(Å)\). A sodium lamp sends incident plane wave onto silt of width \(2\) micrometre. If the screen is located \(2\,\rm{m}\) from the silt, then what is the spacing between the first maxima of two sodium lines as measured on the screen.*

** Ans:** Given: \(λ_1 = 5890\,(Å) = 5890 \times 10^{-10}\,\rm{m}\)

\(λ_2 = 5896 (Å) = 5896 \times 10^{-10}\,\rm{m}\)

\(a = 2\,\rm{μm} = 2 \times 10^{-6}\,\rm{m}\)

\(D = 2\,\rm{m}\)

Now, for the first maxima,

We have,

\(\sin \theta = \frac{{3{\lambda _1}}}{{2a}} = \frac{{{x_1}}}{D}\)

\( \Rightarrow {x_1} = \frac{{3{\lambda _1}D}}{{2a}}\)

And,

\( \Rightarrow {x_2} = \frac{{3{\lambda _2}D}}{{2a}}\)

Therefore, spacing between the first maxima of two sodium lines \(= x_2 – x_1\)

\(\frac{{3D}}{{2a}}\left({{\lambda _2} – {\lambda _1}} \right)\)

\( \Rightarrow \frac{{3 \times 2}}{{2 \times 2 \times {{10}^{ – 6}}}}(5896 – 5890) \times {10^{ – 10}}\)

\( \Rightarrow {x_2} – {x_1} = 9 \times {10^{ – 4}}{\text{m}}\)

Hence, spacing between the first maxima of two sodium lines is \(9 \times 10^{-4}\,\rm{m}\).

Light does not always travel in a straight path; it bends around the corners of an obstacle into the region of the geometrical shadow of the obstacle. While Diffraction occurs whenever propagating waves encounter such changes, its effects are generally most pronounced for waves whose wavelength is roughly equivalent to the size of the diffracting object or slit, and the incident light is from a coherent source.

During Diffraction, it sends a wave around the edges of an opening or other obstacle, and then a single slit produces an interference pattern characterized by a broad central maximum, and also dimmer maxima appears to the edges. The minima will form when, \(\sin \theta = m\frac{\lambda }{a}\). Where \(a\) is the width of the slit, \(λ\) is the wavelength of light and \(m = ±1,\,±2,\,±3\)… Maxima and minima will form in alternate order. The intensity of central maxima will be maximum, and the intensity decreases rapidly on either side.

* Q.1. How can we see Diffraction in our daily life?Ans:* The effects of Diffraction are usually seen in everyday life. For example, when you observe a CD or DVD, the closely spaced tracks on a CD or DVD act as a diffraction grating to form the familiar rainbow pattern. The hologram on a credit card also works with the principle of Diffraction.

* Q.2. Is a rainbow an example of Diffraction?Ans:* No, the rainbow is not an example of Diffraction because a rainbow is formed due to reflection and refraction. This Diffraction does not even play any role in the formation of a rainbow.

* Q.3. What is the application of Diffraction of light?* i. X-ray diffraction is used in crystallography.

Ans:

ii. Diffraction Grating is used to transform the light into its spectrum, and

iii. It is also used in holography.

** Q.4. What is the difference between Diffraction and interference?** The crucial difference between Diffraction and interference is that Diffraction of light occurs due to the superposition of secondary wavelets that generates from various parts of a wavefront. But interference is the result of the superposition of light waves from two coherent sources.

Ans:

*Q.5. What is diffraction grating?Ans:*

*We hope this detailed article on the Diffraction of Light helps you prepare. If you get stuck, do let us know in the comments section below, and we will get back to you at the earliest.*