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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
The intensity at the central maxima in Young’s double slit experiment is I0. Find out the intensity at a point where the path difference is` lambda/6,lambda/4 and lambda/3.`
What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?
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.
If Young's double slit experiment is performed in water, _________________ .
Two transparent slabs having equal thickness but different refractive indices µ1 and µ2are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P0 which is equidistant from the slits?
White coherent light (400 nm-700 nm) is sent through the slits of a Young's double slit experiment (see the following figure). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?
In Young’s double-slit experiment, using monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 x 10-2 m towards the slits, the change in the fringe width is 3 x 10-5 m. If the distance between the two slits is 10-3 m, calculate the wavelength of the light used.
When a beam of light is used to determine the position of an object, the maximum accuracy is achieved, if the light is ______.
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).
In Young's double slit experiment, show that:
`β = (λ"D")/"d"`
Where the terms have their usual meaning.
