- It is a common observation with the waves of all kind that they bend round the edge of an obstacle

- Light like other waves also bends round corners but in comparison to sound waves small bending of light is due to very short wavelength of light which is of the order of 10
^{-5}

- This effect of bending of beams round the corner was first discovered by grimed (Italy 1618-1663)

- We now define diffraction of light as the phenomenon of bending of light waves around the corners and their spreading into the geometrical shadows

- Fresnel then explained that the diffraction phenomenon was the result of mutual interference between the secondary wavelets from the same dif wave front

- Thus we can explain diffraction phenomenon using Huygens’s principle

- The diffraction phenomenon are usually divided into two classes

i) Fresnel class of diffraction phenomenon where the source of light and screen are in general at a finite distance from the diffracting aperture

ii) Fruanhofer class of diffraction phenomenon where the source and the screen are at infinite distance from the aperture, this is easily achieved by placing the source on the focal plane of a convex lens and placing screen on focal plane of another convex lens. This class of diffraction is simple to treat and easy to observe in practice

- Here in this chapter we will only be considering fraunhofer class diffraction by a single slit

- Let us first consider a parallel beam of light incident normally on a slit AB of width 'a' which is of order of the wavelength of light as shown below in the figure

- A real image of diffraction pattern is formed on the screen with the help of converging lens placed in the path of the diffracted beam

- All the rays that starts from slit AB in the same phase reinforce each other and produce brightness at point O on the axis of slit as they arrive there in the same phase

- The intensity of diffracted beam will be different in different directions and there are some directories where there is no light

- Thus diffraction pattern on screen consists of a central bright band and alternate dark and bright bands of decreasing intensity on both sides

- Now consider a plane wave front PQ incident on the narrow slit AB. According to Huygens principle each point t on unblocked portion of wavefront PQ sends out secondary wavelets in all directions

- Their combined effect at any distant point can be found y summing the numerous waves arriving there from the principle of superposition

- Let C be the center of the slit AB.The secondary waves, from points equidistant from center C of the clit lying on portion CA and CB of wave front travel the same distance in reaching O and hence the path difference between them is zero

- These waves reinforce each other and give rise to the central maximum at point O

- We now consider the intensity at point P
_{1}above O on the screen where another set of rays diffracted at a angle θ have been bought to focus by the lens and contributions from different elements of the slits do not arise in phase at P_{1}

- If we drop a perpendicular from point A to the diffracted ray from B,then AE as shown in figure constitutes the diffracted wavefront and BE is the path difference between the rays from the two edges A and B of the slit.

- Let us imagine this path difference to be equal to one wavelength.

- The wavelets from different parts of the slit do not reach point P
_{1}in the phase because they cover unequal distance in reaching P_{1}.Thus they would interfere and cancel out each other effect. For this to occur

BE=λ

Since BE=ABsinθ

asinθ=λ

or sinθ=λ/a

or θ=λ/a ---(1)

As angle of diffraction is usually very small so that

sinθ=θ

- Such a point on screen as given by the equation (1) would be point of secondary minimum

- It is because we have assume the slit to be divided into two parts, then wavelets from the corresponding points of the two halves of the slit will have path difference of #955;/2 and wavelets from two halves will reach point P
_{1}on the screen in a opposite phase to produce minima

- Again consider the point P
_{2}in the figure 1 and if for this point path difference BE=2λ ,then we can imagine slit to be divided into four equal parts

- The wavelets from the corresponding points of the two adjacent parts of the slit will have a path difference of λ/2 and will mutually interfere to cancel out each other

- Thus a second minimum occurs at P
_{2}in direction of θ given by

θ=2θ/a

- Similarly n
_{th}minimum at point P_{n}occurs in direction of θ given by

θ_{n}=nθ/a ---(2)

- If there is any point on the screen for which path difference

BN=asinθ=3θ/2

Then point will be position of first secondary maxima

- Here we imagine unblocked wavefront to be divided into three equal parts where the wavelets from the first two parts reach point P in opposite phase thereby cancelling the e effects of each other

- The secondary waves from third part remain uncancelled and produce first maximum at the given point

- we will get second secondary maximum for BN=5θ/2 and n
_{th}secondary maxima for

BN=(2n+1)θ/2 =asinθ_{n}---(3)

where n=1,2,3,4..

- Intensity of these secondary maxima is much less then central maxima and falls off rapidly as move outwards

- Figure below shows the variation of the intensity distribution with their distance from the center of the central maxima

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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