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प्रश्न
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.
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उत्तर
Young's experiment to show interference of light passing through two slits. A pattern of bright and dark areas appears on the screen (as shown in figure (i)).
The condition for destructive interference is `Δx = S_2P - S_1P +- ((2n - 1)/2)lambda` where n = 1, 2, ...
For nth minima to be formed on the screen path difference (Δx) between the rays coming from S1 and S2 must be `((2n - 1)/2)lambda`
The minima will occur when `Δx = S_2P - S_1P = ((2n - 1)/2)lambda` ......(i)
From the given figure,
`S_1P = sqrt((S_1T_1)^2 + (PT_1)^2) = sqrt(D^2 + (D - x)^2)`
And `S_2P = sqrt((S_2T_2)^2 + (T_2P^2)) = sqrt(D^2 + (D + x)^2)`
`T_2P = T_2O + OP = D + x`
And `T_1P = T_1O - OP = D - x`
Hence, `[D^2 + (D + x)^2]^(-1/2) - [D^2 + (D - x)^2]^(1/2) = lambda/2` ......[For first minima n = 1]
If `x = D`
We can write, `[D^2 + 4D^2]^(-1/2) - [D^2 + 0]^(1/2) = lambda/2`
⇒ `[5D^2]^(1/2) - [D^2 + 0]^(1/2) = lambda/2`
⇒ `sqrt(5)D - D = lambda/2` or `D = lambda/(2sqrt(5) - 1)` = 0.404 λ
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संबंधित 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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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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In Young's double slit experiment the slits are 0.589 mm apart and the interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 7.5 mm from the dark fringe which is second from the center of the fringe pattern. Find the wavelength of the light used.
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 ______.
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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- the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
- the least distance from the central maximum where the bright fringes due to both wavelengths coincide.
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- Slits are brought closer to each other.
- Screen is moved away from the slits.
- Red coloured light is replaced with blue coloured light.
