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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 Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
The fringes produced in diffraction pattern are of _______.
(A) equal width with same intensity
(B) unequal width with varying intensity
(C) equal intensity\
(D) equal width with varying intensity
The intensity at the central maxima in Young’s double slit experimental set-up is I0. Show that the intensity at a point where the path difference is λ/3 is I0/4.
A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?
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) \[\frac{2D\lambda}{d}\]
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 ______.
How will the interference pattern in Young's double-slit experiment be affected if the phase difference between the light waves emanating from the two slits S1 and S2 changes from 0 to π and remains constant?
- 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, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
