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प्रश्न
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.
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उत्तर
From the figure,
Δx = AMO − ANO
Here AM = MO = √[D2 + d2] AND AN = NO = D
=> Δx = 2(AM − AN) = 2 {√[D2 + d2] - D}
For minima at O,
2 {√[D2 + d2] − D} = (n +(1/2)) λ
Solving above equation for d, we get
d = √[Dλ/2]
(b) width of the dark fringe = w = Dλ/d
Now, the location x is given by
x = Dλ/[2√(Dλ/2)]
=> x = d
(c) As x = w/2
=> w = 2x = 2d
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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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Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
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(i) constructive interference and
(ii) destructive interference at a point on the screen.
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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(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(c) The fringe next to the central will be red.
(d) The fringe next to the central will be violet.
A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light \[\left( \lambda = 700\text{ nm in vacuum} \right).\] Find the fringe-width of the pattern formed 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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