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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
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
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?
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) \[\frac{2D\lambda}{d}\]
Write the conditions on path difference under which constructive interference occurs in Young’s double-slit experiment.
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, 4mm apart, are illuminated by light of wavelength 6000 A° what will be the fringe width on a screen placed 2 m from the slits?
A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit. Find:
- The distance of the second bright fringe from the central maximum pertaining to the light of 600 nm.
- The least distance from the central maximum at which bright fringes due to both wavelengths coincide.
- 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.
