Advertisements
Advertisements
प्रश्न
Explain Fraunhofer diffraction at a single slit with a neat ray diagram. Obtain an expression for the width of the central bright fringe.
उत्तर
Fraunhofer diffraction due to single slit:
- Consider a narrow slit AB of width ‘a’, kept perpendicular to the plane of the paper. The slit can be imagined to be divided into extremely thin slits or slit elements. It is illuminated by a parallel beam of monochromatic light of wavelength λ i.e., a plane wavefront is an incident on AB.
- The diffracted light is focused by a converging lens L, on a screen XY.
- The screen is kept in the focal plane of the lens and is perpendicular to the plane of the paper.
- Let D be the distance between the slit and the screen.
- According to Huygens’ principle, each and every point of the slit acts as a source of secondary wavelets, spreading in all directions.
Position of central maxima:
- Let C be the centre of the slit AB. The secondary wavelets traveling parallel to CPo come to a focus at Po. The secondary wavelets from points equidistant from C in the upper and lower halves of the slit travel equal paths before reaching Po.
- The optical path difference between all these wavelets is zero and hence they interfere in the same phase forming a bright image at Po.
- The intensity of light is maximum at the point Po. It is called the central or the principal maxima of the diffraction pattern.
- For the line CPo, angle θ = 0°.
Position of secondary minima:
- Consider a point P on the screen at which waves travelling in a direction making an angle θ with CP are brought to focus at P by the lens. This point P will be of maximum or minimum intensity because the waves reaching P will cover the unequal distance.
- Draw AN perpendicular to the direction of diffracted rays from point A. BN is the path difference between secondary waves coming from A and B.
- From ΔABN, sin θ = `"BN"/"AB"`
∴ BN = AB sin θ = a sin θ
Since θ is very small
∴ sin θ ≈ θ
∴ BN = aθ
In figure, suppose BN = λ and θ = θ1 then sin θ1 = `lambda/"a"` - Such a point on the screen will be the position of the first secondary minimum. It is because, if the slit is assumed to be divided into two equal halves AC and BC, then the waves from corresponding points of two halves of the slit will have a path difference of λ/2.
It gives rise to destructive interference at P which has minimum intensity. - If point P on the screen is such that BN = 2λ and angle θ = θ2, then, sin θ2 = `(2λ)/"a"`. Such a point on the screen will be the position of the second secondary minimum.
In general, for nth minimum, sinθn = `("n"lambda)/"a"`
where, n = ±1, ±2, ±3,… - If ynd is the distance of nth minimum from Po, on the screen, then `(tan theta_"nd") = ("y"_"nd")/"D"`.
- If θnd is very small,
- tan θnd ≈ sin θnd = `("n"lambda)/"a"`
- ∴ `("y"_"nd")/"D" = ("n"lambda)/"a"`
- ∴ `"y"_"nd" = ("n"lambda"D")/"a"` = nW ....(1)
where, W is fringe width Equation (1) gives distance of nth secondary minima from central maxima. - The central bright fringe is spread between the first dark fringes on either side. Hence, width of the central bright fringe is the distance between the centres of first dark fringe on either side.
∴ Width of the central bright fringe,
Wc = 2y1d = 2W = 2`((lambda"D")/"a")`
This is the required expression for the width of the central bright fringe.
APPEARS IN
संबंधित प्रश्न
In Fraunhoffer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light.
Explain experimental setup for Fraunhofer diffraction with neat diagram.
A lens having focal length f gives Fraunhofer type diffraction pattern of a slit having width a. If wavelength of light is λ, the distance of first dark band and next bright band from axis is given by ____________.
Light of wavelength 'λ' is incident on a single slit of width 'a' and the distance between slit and screen is 'D'. In diffraction pattern, if slit width is equal to the width of the central maximum then 'D' is equal to ______.
A slit of width a is illuminated by white light. For red light `(λ = 6500 Å)`, the first minima is obtained at θ = 60°. Then the value of a will be ______.
For Fraunhofer diffraction to occur ____________.
In a diffraction pattern, width of a fringe ____________.
In a single slit diffraction pattern, which of the following is incorrect for fringe pattern?
A slit of width 'a' illuminated by white light. The first diffraction minimum for light of wavelength 6500 Å is formed at θ = 30°, then 'a' is (sin 30° = 0.5).
The condition for observing Fraunhofer diffraction pattern from an obstacle is that the light wavefront incident on it must be ______.
Select the CORRECT statement from the following.
In Fresnel's biprism experiment, when the distance between the slit aperture and eye is increased, then distance between the fringes ____________.
The first minimum of a single slit diffraction pattern is observed at angle 2° with a light of wavelength 510 nm. The width of this slit is ____________.
In Fraunhofer diffraction due to single slit, the angular width of central maximum does 'NOT' depend on ______
In young 's double slit experiment the two coherent sources have different amplitudes. If the ratio of maximum intensity to minimum intensity is 16 : 1, then the ratio of amplitudes of the two source will be _______.
A physical quantity P is described by the relation p = a1/2 b2 c3 d-4. If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be ______.
In a Young's double-slit experiment, let β be the fringe width, and let I0 be the intensity at the central bright fringe. At a distance x from the central bright fringe, the intensity will be ______.
The increase in energy of a metal bar of length 'L' and cross-sectional area 'A' when compressed with a load 'M' along its length is ______.
(Y = Young's modulus of the material of metal bar)
In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 Å, then the distance between the first minimum on either side of the central maximum is ______. (θ is small and measured in radian)
A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping intererence patterns on the screen. The central maximum of both lights coincides.
Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data the wavelength of the unknown light is ______.
In Young's double slit experiment, which of the following graph represents the correct variation of fringe width β versus distance D between sources and screen?
In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 µ-m and a width of 4.05 µm. The number of bright fringes between the first and the second diffraction minima is ______.
In a double slit interference experiment, the distance between the slits is 0.05 cm and screen is 2 m away from the slits. The wavelength of light is 6000 Å. The distance between the fringes is ______.
State the characteristics of a single slit diffraction pattern.
