• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer

CBSE Tuts

CBSE Maths notes, CBSE physics notes, CBSE chemistry notes

  • NCERT Solutions
    • NCERT Solutions for Class 12 English Flamingo and Vistas
    • NCERT Solutions for Class 11 English
    • NCERT Solutions for Class 11 Hindi
    • NCERT Solutions for Class 12 Hindi
    • NCERT Books Free Download
  • TS Grewal
    • TS Grewal Class 12 Accountancy Solutions
    • TS Grewal Class 11 Accountancy Solutions
  • CBSE Sample Papers
  • NCERT Exemplar Problems
  • English Grammar
    • Wordfeud Cheat
  • MCQ Questions

Classification of Diffraction Phenomena – Fraunhofer Diffraction By Single Slit

Contents

The laws of Physics Topics are used to explain everything from the smallest subatomic particles to the largest galaxies.

What are the Similarities Between Diffraction and Interference of Light?

Similarity: Both interference and diffraction of light take place due to superposition of waves. Diffraction fringes are formed mainly due to the interference of waves.

Dissimilarity: There are some basic differences between interference and diffraction of light. The differences are as follows:

Interference Diffraction
1. Interference is due to superposition of light waves coming from two coherent sources. 1. Diffraction is due to interference of secondary wavelets coming from the different points of the same wavefront.
2. Interference fringes generally have same width. 2. Diffraction fringes never have same width.
3. All bright fringes in interference pattern have uniform intensity. 3. Not all bright fringes in diffraction pattern have same intensity.
4. Dark bands of the interference fringes are perfectly dark. 4. Dark bands of the diffraction fringes are not perfectly dark. They are partially dark.

Classification of Diffraction

The phenomena of diffraction of light can be classified mainly into two classes—

  1. Fresnel diffraction and
  2. Fraunhofer diffraction.

Fresnel diffraction: The diffraction, where both the source of light and the screen, are at finite distances from the obstacle or the aperture is called Fresnel diffraction. In this diffraction, wavefronts incident on screen are either spherical or cylindrical.

Obstacles with sharp edges, narrow slit, thin wire, small circular obstacle or hole etc. can produce Presnel diffraction.

Fraunhofer diffraction: The diffraction, where the source of light and the screen are virtually at infinite distance, is called Fraunhofer diffraction. In this case, the incident wave-front is plane.

Single slit, double slit, diffraction grating etc. produce Fraunhofer diffraction.

Fraunhofer Diffraction By Single Slit

Experimental arrangement: In this experiment, a ray of light from a monochromatic light source, is made to fall on a convex lens L1 through a narrow slit S. The slit S is held at the focus of the lens L1. Hence, rays refracted from lens L1 become parallel. [Fig.].
Classification of Diffraction Phenomena Fraunhofer Diffraction By Single Slit 1
This parallel beam of monochromatic light is incidênt normally on the slit AB placed perpendicular to the plane of the paper. The ray is now focussed by a convex lens L2 on a screen MN, where we observe diffraction fringes. [Fig.]. Instead of the screen, if the diffraction pattern is observed by an objective, the pattern will be observed in its focal plane.
Classification of Diffraction Phenomena Fraunhofer Diffraction By Single Slit 2

Explanation: According to geometrical optics, light rays emerging out from the slit AB, if focussed by the lens L2, should produce a sharp image of the slit, at point O of the screen. But actually this does not happen. This is because light rays passing through AB, do not propagate in straight lines. Getting diffracted by AB, the light rays spread upwards of point A and downwards of point B. So with the formation of a sharp image of the slit at O, alternate bright and dark diffraction fringes are produced on both sides of O [Fig.].

Central or principal maximum: As shown in Fig., C is the midpoint of the slit AB. CO is the principal axis. Every point of the plane wavefront, which is incident on the slit, is of the same phase. All wavelets originating from those points and proceeding parallel to CO are focussed by L2 at O. Since these wavelets have no path difference, they are in the same phase. So, they make constructive interference and the point O appears very bright, O is called principal or central maximum. Simplified form of Fig. is shown in Fig.
Classification of Diffraction Phenomena Fraunhofer Diffraction By Single Slit 3
Conditions for the formation of minima and secondary maxima: Suppose, some wavelets after being diffracted through an angle θ are focussed at O1 by lens L2. Now, the condition for formation of a constructive or a destructive interference at the point O1 depends on the path difference of the wavelets originating from A and B.

From A, a perpendicular AP is drawn on BO1. So the path difference between the wavelets emergent from points A and B = BP.
Now, BP = ABsin∠BAP = a sinθ
[where, AB = a = width of the slit]

For minima: To obtain the condition for minima being formed at O1, slit AB is notionally divided into two equal halves, AC and CB,i.e., AC = CB = \(\frac{a}{2}\).

Let the wavelength of incident monochromatic light = λ. If the path difference between two wavelets, originating from points A and C be \(\frac{\lambda}{2}\), they would cause destructive interference.

So, the condition of formation of first minima on both sides of O for diffracting angle θ1 is,
\(\frac{a}{2}\)sinθ1 = \(\frac{\lambda}{2}\) or, a sin θ1 = λ
or, sin θ1 = \(\frac{\lambda}{a}\) …. (1)

In general, the condition for the formation of n th minimum on both sides of O, for diffracting angle θn is,
a sinθn = nλ ….. (2)

Putting n = ±1, ±2, ±3, in equation (2), we would get simultaneously, first, second, third etc. minima on either side of the principal maximum. Here, ± sign is used to indicate diffractions on either side of the central line.

For secondary maxima: If the path difference of the wavelets emitted from A and B, BP = \(\frac{3 \lambda}{2}\), \(\frac{5 \lambda}{2}\), …… (2n + 1)\(\frac{\lambda}{2}\) then at points O2, O4 ….. etc. they would produce first, second, etc. maxima. At these points, the two waves superpose in the same phase. These are called secondary maxima.

If θ’n be the corresponding angle of diffraction for the n th secondary maximum, then
a sinθ’n = (2n + 1)\(\frac{\lambda}{2}\) ….. (3)
[where, n = ±1, ±2, ±3….. etc.]
It is to be noted that the intensity of the secondary maxima gradually decreases [see Fig.].

Linear distance of nth minimum from central maximum: Generally, the wavelength of visible light (for example, λ = 5 × 10-7m) is much lower than the width of the slit (for example, a = 10-4m). For such values of θ, sin θ ≈ θ. With this approximation, equation (2) becomes,
θn = \(\frac{n \lambda}{a}\) …. (4)

Let the distance from principal maximum point O to n th minimum point On, OOn = xn and distance from screen to slit = D.
As the value of θn, is very small, θn = \(\frac{x_n}{D}\).
Putting the value of θn, in equation (4) we get,
a ᐧ \(\frac{x_n}{D}\) = nλ or, xn = \(\frac{n \lambda D}{a}\) ……. (5)
Putting n = ±1, ±2, ±3…….. etc. in equation (5), linear distances of various minima from central maximum are obtained.

Width of central maximum: The angle between the first minima on either side of central maximum is called the angular width of central maximum.

According to equation (4), the angular spread of the central maximum on either side is,
θ = \(\frac{\lambda}{a}\)
∴ Angular width of central maximum,
2θ = \(\frac{2 \lambda}{a}\) ….. (6)
Therefore, linear width of central maximum D ᐧ 2θ = \(\frac{2 D \lambda}{a}\), where D = distance of slit from the screen.

If lens L2 is located vent close to the slit AB, or if the screen MN is located far away from lens L2, then D ≈ focal length of the lens (f). In that case, linear width of central maximum point = \(\frac{2 f \lambda}{a}\).

1. If the silt is very narrow for single slit Fraunhofer diffraction, the condition of formation of first minima on either side of the central maximum for diffraction angle θ is,
a sinθ = ±λ or, sinθ = ±\(\frac{\lambda}{a}\)

Now, if the value of a is decreased, the value of θ increases i.e., the central maximum becomes wider. When value of a is equal to λ, the first minima is formed at θ = 90° i.e., central maximum occupies the whole area.

2. If the slit is illuminated by white light for single slit Fraunhofer diffraction, the central maximum band will be white, but the secondary maxima bands will be coloured.

Primary Sidebar

NCERT Exemplar problems With Solutions CBSE Previous Year Questions with Solutoins CBSE Sample Papers
  • The Summer Of The Beautiful White Horse Answers
  • Job Application Letter class 12 Samples
  • Science Lab Manual Class 9
  • Letter to The Editor Class 12 Samples
  • Unseen Passage For Class 6 Answers
  • NCERT Solutions for Class 12 Hindi Core
  • Invitation and Replies Class 12 Examples
  • Advertisement Writing Class 11 Examples
  • Lab Manual Class 10 Science

Recent Posts

  • Understanding Diversity Question Answer Class 6 Social Science Civics Chapter 1 NCERT Solutions
  • Our Changing Earth Question Answer Class 7 Social Science Geography Chapter 3 NCERT Solutions
  • Inside Our Earth Question Answer Class 7 Social Science Geography Chapter 2 NCERT Solutions
  • Rulers and Buildings Question Answer Class 7 Social Science History Chapter 5 NCERT Solutions
  • On Equality Question Answer Class 7 Social Science Civics Chapter 1 NCERT Solutions
  • Role of the Government in Health Question Answer Class 7 Social Science Civics Chapter 2 NCERT Solutions
  • Vital Villages, Thriving Towns Question Answer Class 6 Social Science History Chapter 9 NCERT Solutions
  • New Empires and Kingdoms Question Answer Class 6 Social Science History Chapter 11 NCERT Solutions
  • The Delhi Sultans Question Answer Class 7 Social Science History Chapter 3 NCERT Solutions
  • The Mughal Empire Question Answer Class 7 Social Science History Chapter 4 NCERT Solutions
  • India: Climate Vegetation and Wildlife Question Answer Class 6 Social Science Geography Chapter 8 NCERT Solutions
  • Traders, Kings and Pilgrims Question Answer Class 6 Social Science History Chapter 10 NCERT Solutions
  • Environment Question Answer Class 7 Social Science Geography Chapter 1 NCERT Solutions
  • Understanding Advertising Question Answer Class 7 Social Science Civics Chapter 7 NCERT Solutions
  • The Making of Regional Cultures Question Answer Class 7 Social Science History Chapter 9 NCERT Solutions

Footer

Maths NCERT Solutions

NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 11 Maths
NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 9 Maths
NCERT Solutions for Class 8 Maths
NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 6 Maths

SCIENCE NCERT SOLUTIONS

NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 11 Physics
NCERT Solutions for Class 11 Chemistry
NCERT Solutions for Class 10 Science
NCERT Solutions for Class 9 Science
NCERT Solutions for Class 7 Science
MCQ Questions NCERT Solutions
CBSE Sample Papers
NCERT Exemplar Solutions LCM and GCF Calculator
TS Grewal Accountancy Class 12 Solutions
TS Grewal Accountancy Class 11 Solutions