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प्रश्न
In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
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उत्तर
In Young's double-slit experiment, the wavefronts from the two illuminated slits superpose on the screen. This leads to formation of alternate dark and bright fringes due to constructive and destructive interference, respectively. At the centre C of the screen, the intensity of light is maximum and it is called central maxima.
Let S1 and S2 be two slits separated by a distance d. GG'is the screen at a distance D from the slits S1 and S2. Point C is equidistant from both the slits. The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits S1 and S2 is S2P - S1P.
Now, S1 S2 = d, EF = d, and S2F= D
∴In ΔS2PF,
`S_2P=[S_2F^2+PF^2]^(1/2)`
`S_2P=[D^2+(x+d/2)^2]^(1/2)`
`=D[1+(x+d/2)/D^2]^(1/2)`
Similarly, in ΔS1PE,
`S_1P=D[1+(x-d/2)^2/D^2]^(1/2)`
`:.S_2P-S_1P=D[1+1/2(x+d/2)^2/D^2]-D[1+1/2(x-d/2)^2/D^2]`
On expanding it binomially,
`S_2P-S_1P=1/(2D)[4xd/2]=(xd)/D`
For bright fringes (constructive interference), the path difference is an integral multiple of wavelengths, i.e. path difference is nλ.
`:.nlambda=(xd)/D`
`x=(nlambdaD)/d`where n = 0, 1, 2, 3, 4, …
For n = 0, x0 = 0
n =1, `x_1=(lambdaD)d`
n = 2, `x_2=(2lambdaD)/d`
n =3, `x_3=(3lambdaD)d`
`n=n, x_n = (nlambdaD)/d`
Fringe width (β) → Separation between the centres of two consecutive bright fringes is called the width of a dark fringe.
`:.beta_1=x_n-x_(n-1)=(lambdaD)/d`
Similarly, for dark fringes,
`x_n=(2n-1)lambda/2D/d`
For n =1, `x_1=(lambdaD)/(2d)`
For n =2, `x_2=(3lambdaD)/(2d)`
The separation between the centres of two consecutive dark interference fringes is the width of a bright fringe.
`:. beta_2=x_n-x_(n-1)=(lambdaD)/d`
∴β1 = β2
All the bright and dark fringes are of equal width as β1 = β2
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संबंधित प्रश्न
Write two characteristics features distinguish the diffractions pattern from the interference fringes obtained in Young’s double slit experiment.
A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
In Young's double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e. fringe separation.
"If the slits in Young's double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit".
Justify the above statement through a relevant mathematical expression.
In Young's double slit experiment, the minimum amplitude is obtained when the phase difference of super-imposing waves is: (where n = 1, 2, 3, ...)
In Young's double slit experiment shown in figure S1 and S2 are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has strong intensity?
In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case ______.
Consider a two-slit interference arrangement (Figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.
- Assertion (A): In Young's double slit experiment all fringes are of equal width.
- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
In Young’s double slit experiment, how is interference pattern affected when the following changes are made:
- Slits are brought closer to each other.
- Screen is moved away from the slits.
- Red coloured light is replaced with blue coloured light.
