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प्रश्न
Derive an expression for path difference in Young’s double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
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उत्तर
Diagrammatical arrangement for Young’s double slit experiment:
In the above figure, S1 and S2 are two narrow closely spaced slits illuminated by monochromatic light of wavelength λ. The screen on which the interference pattern is observed is represented as XY.
If S1 and S2 emit light in the same phase, then for point O, the path difference receives light in the same phase. The superposition at O is constructive producing a bright point called the central maxima. The intensity at any point P at a distance x from O depends on the path difference between light reaching P from S1 and S2. The path difference is S2P − S1P.
From the geometry of the figure, we have
`(S_2P)^2=D^2+(x+d/2)^2`
Similarly, we have
`(S_1P)^2=D^2+(x+d/2)^2`
`:.(S_2P)^2-(S_1P)^2=D^2+(x+d/2)^2-D^2-(x-d/2)^2`
`:.(S_2P)^2-(S_1P)^2=(x+d/2)^2-(x-d/2)^2`
`:.(S_2P)^2-(S_1P)^2=x^2+xd+d^2/4-x^2+xd-d^2/4`
∴ (S2P)2-(S1P)2=2xd
∴(S2P-S1P)(S2P+S1P)=2xd
`:.S_2P-S_1P=(2xd)/(S_2P+S_1P)`
Now, from the figure we can see that
S2P≈S1P = D
`:.S_2P-S_1P=(2xd)/(2D)=(xd)/D`
This is the expression for path difference.
Condition for constructive interference: Constructive interference will occur when the phase difference between the two superposing waves is an even multiple of π or the path difference is an integral multiple of wavelength λ.
Condition for destructive interference: Destructive interference will occur when the phase difference between the two superposing waves is an odd multiple of π or the path difference is an odd multiple of wavelength λ/2.
संबंधित प्रश्न
(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.
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ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
REASON (R): Fringe width is proportional to (d/D).
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Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
Two beams of light having intensities I and 41 interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xl. The value of x will be ______.
In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.
- 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, show that:
`β = (λ"D")/"d"`
Where the terms have their usual meaning.
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.
