Advertisements
Advertisements
प्रश्न
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
Advertisements
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 young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.
Show that the fringe pattern on the screen is actually a superposition of slit diffraction from each slit.
How does an unpolarized light incident on a polaroid get polarized? Describe briefly, with the help of a necessary diagram, the polarization of light by reflection from a transparent medium.
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.
White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe \[\left( \lambda = 400\text{ nm} \right)\] which overlaps with a red fringe \[\left( \lambda = 700\text{ nm} \right).\]
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.
In a Young's double slit experiment, \[\lambda = 500\text{ nm, d = 1.0 mm and D = 1.0 m.}\] Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
When a beam of light is used to determine the position of an object, the maximum accuracy is achieved, if the light is ______.
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit- width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :
