Class 12 Physics Wave Optics Fringe Pattern

Fringe Pattern

• The alternate dark and red bands which are obtained on the screen are known as fringe pattern and the alternate dark and bright bands are known as fringes.
• Bright Bands:-
• Bright bands are formed as a result of constructive interference and they are the positions of maximum intensity.
• Condition for maximum intensity:-
• Path difference =n λ.
• =>(xd/D) = nλ using Equation(2)
• =>xn = ((n λ D)/d) where xn=position of nth bright band.
• When n=0 then it will be central bright band.
• Dark Bands:-
• Dark bands are formed by the destructive interference and they are the positions of minimum intensity.
• Condition for destructive interference:-
• Path difference =(n+(1/2))λ
• => (xd/D) = (n+ (1/2)) λ.
• =>xn=(n+ (1/2)) (λD/d)
• Where xn=position of nth dark band.

Graphical representation of fringe pattern

Fringe width:-

1. Fringe width is the distance between consecutive dark and bright fringes.
2. It is denoted by ‘β’.
3. In case of constructive interference fringe width remains constant throughout.
4. It is also known as linear fringe width.

Angular Fringe width:-

1. It is the angle subtended by a dark or bright fringe at the centre of the 2 slits.
2. It is denoted by ‘θ’.
• Mathematical Expression for fringe width(β):-
• xn =((nλD)/d)
• xn+1 =(((n+1)λD)/d)
• β =xn+1 - xn
• =(((n+1) λD)/d) – ((nλD)/d)
• =>β = (λD/d)
• Therefore fringe width depends on:-
• (λ)Wavelength of the light used, (D) distance of the screen from the slits and (d) distance between two slits.
• Mathematical Expression forangular fringe width(θ):-
• θ =(β/D)
• θ = (λD)/(d D)
• θ = (λ/d)

Conclusion of Young’s double slit experiment

1. Central fringes gets shifted by – θ if the source gets shifted by θ.
1. If the source S is shifted by some angle θ, there will be no change in fringe pattern. The central fringe will get shifted in the opposite direction.
2. Intensity of the fringes increase if point sources are replaced with slits.
1. If there are slits instead of point source then more light waves will be able to pass through the slits.
2. As a result stronger wavefronts are formedwhich give rise to even greater intensity fringes.

Problem:- In a Young’s double-slit experiment, the slits are separated by0.28 mm and the screen is placed 1.4 m away. The distance betweenthe central bright fringe and the fourth bright fringe is measuredto be 1.2 cm. Determine the wavelength of light used in theexperiment.

Answer:- Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m

Distance between the slits and the screen, D = 1.4 m

Distance between the central fringe and the fourth (n = 4) fringe,

u = 1.2 cm = 1.2 × 10−2 m

In case of a constructive interference, we have the relation for the distance between thetwo fringes as:

u =(n λD)/(d)

Where, n = Order of fringes

= 4 λ = Wavelength of light used

Therefore, λ= (ud/nD)

= (1.2x10-2x0.28x10-3)/ (4x1.4)

=6x10-7

=600nm

Hence, the wavelength of the light is 600 nm.

Problem:- In Young’s double-slit experiment using monochromatic light of wavelengthλ, theintensity of light at a point on the screen where path difference is λ, is K units. What isthe intensity of light at a point where path difference is λ /3?

Answer:- Let I1 and I2 be the intensity of the two light waves. Their resultant intensities can be I’ = I1 + I2 +2√ I1 I2 cosφobtained as:

Where,

φ = Phase difference between the two waves

For monochromatic light waves,

I1=I2

Therefore I’ = I1 + I2 +2√ I1 I2 cosφ

=2 I1 + 2I1cosφ

Phase difference =(2π/λ) x Path difference

Since path difference = λ,

Phase difference,φ =2π

Therefore, I’ =2 I1 + 2I1 = 4I1

Given,

I’ = K

Therefore, I1 = (k/4)   (equation (1))

When path difference= λ /3,

Phase differenceφ = (2π/3),

Hence, resultant intensity,

IR =I1 + I1+2√ I1 I2 cos(2π/3),

=2I1+2I1(-1/2) =I1

Using equation (1), we can write:

IR = I1 = (k/4)

Hence, the intensity of light at a point where the path difference is λ /3 units is (k/4) units.

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