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प्रश्न
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
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उत्तर
Young’s double slit experiment: Consider two narrow rectangular slits S1 and S2 placed perpendicular to the plane of paper. Slit S is placed on the perpendicular bisector of S1S2 and is illuminated with monochromatic light.
The slits are separated by a small distance d. A screen is placed at a distance D from S1, S2.
Consider a point P on the screen at distance x from O.
The path difference between the waves reaching P from S1 and S2 is:
P = S2P − S1P
Draw S1N perpendicular to S2P. Then,
P = S2P − S1P = S2P − NP = S2N
From right-angled
`DeltaS_1S^2N=(S_2N)/(S_2S_1) = sintheta`
`therefore P =S_2N=S_2S_1sintheta = d sintheta`
From ΔCOP,
When θ is small,
`sintheta≈theta≈tantheta = x/D`
`therefore P=(xd)/D`
For constructive interference,
`(xd)/D =nlambda,n=0,1,2,3,.....`
Position of nth bright fringe, `x_n = (nDlambda)/d =0,(Dlambda)/d,(2Dlambda)/d,(3Dlambda)/d,.......`
When n = 0, xn = 0, central bright fringe is formed at O.
For destructive interference,
`(xd)/D = (2n +1)lambda/2`
`or x_n = (2_n +1) (lambdaD)/(2d) = 1/2(lambdaD)/d,3/2(lambdaD)/d,5/2(lambdaD)/d,......`
Thus, alternate bright and dark fringes are formed on the screen.
संबंधित प्रश्न
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?
In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength \[\lambda.\] Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.
Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to
(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
(c) \[\frac{2D\lambda}{d}\]
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`β = (λ"D")/"d"`
Where the terms have their usual meaning.
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