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प्रश्न
In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width.
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उत्तर
For any other incoherent source of light a steady interference pattern can never be obtained, even if the sources emit waves of equal wavelengths and equal amplitudes. This is because the waves emitted by a source undergo rapid and irregular changes of phase, so that the intensity at any point is never constant. Naturally the phase difference between the waves emitted by the two sources cannot remain constant.
The two waves interfering at P have different distances S1P = x and S2P = x + Δx.
So, for the two sources S1 and S2we can respectively write,
`I_1 = I_(01) sin (kx -wt)`
`I_1 = I_(02) sin (k(x +Deltax)-wt) =I_(02)sin(kx -wt +delta)`
`delta = kDeltax =((2pi)/lambda) xx Delta x`
The resultant can be written as,
`I =I_0sin(kx -wt +epsi)
Where` I_0^2 =I_(01)^2 + I_(02)^2 +2I_(01) I_(02) cos delta`
And tan`epsi = I_(02) sin delta /(I_(01) +I_(02)cos delta)`
The condition for constructive (bright fringe) and destructive (dark fringe) interference are as follows;
δ = 2nπ for bright fringes
δ = (2n + 1) π for dark fringes
Where n is an integer.
Now to find the fringe width,
The path difference is `Deltax =S_2P-S_1P` nearly equal to d `sintheta =d tantheta =(dy)/D`
Hence we can write, `y=(nlambdaD)/d ,n` is an integer.
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संबंधित प्रश्न
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be ____________ .
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.
Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits.
- Find the minimum value of d so that there is a dark fringe at O.
- Suppose d has this value. Find the distance x at which the next bright fringe is formed.
- Find the fringe-width.
In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is 0.20 W m−2, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes?
In Young’s double slit experiment, what should be the phase difference between the two overlapping waves to obtain 5th dark band/fringe on the screen?
Two slits in Young's interference experiment have width in the ratio 1 : 2. The ratio of intensity at the maxima and minima in their interference is ______.
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 ______.
In a double-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 × 10-2 m towards the slits, the change in fringe width is 3 × 10-3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ______ nm.
What happens to fringe width when slit separation increases while wavelength and screen distance remain unchanged?
