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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
Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.
In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10−7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?
A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is \[\lambda.\] Neglect any absorption of light in the plate.
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.
What should be the path difference between two waves reaching a point for obtaining constructive interference in Young’s Double Slit experiment ?
In Young’s double-slit experiment, show that:
`beta = (lambda "D")/"d"` where the terms have their usual meaning.
In Young’s double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other?
"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 a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is `1/8`th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to ______.
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.
