Advertisements
Advertisements
प्रश्न
(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.
(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.
Advertisements
उत्तर
Expression for Fringe Width in Young’s Double-Slit Experiment
Let S1 and S2 be two slits separated by a distance d. `GG'` is the screen at a distance D from the slits S1and S2. Point C is equidistant from both the slits. The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits is given by
S1= S2P − S1P
Now, S1 S2 = d, EF = d, and S2F = D
In ΔS2PF,
`S_2P=[S_2F^2+PF^2]^(1/2)`
`S_2P=[D2+(x+d/2)^2]^(1/2)`
`=D[1+(x+d/2)^2/D^2]^(1/2)`
Similarly, in ΔS1PE
`S_1P=D[1+(x-d/2)^2/D^2]^(1/2)`
`:.S_2P-S_1P=D[1+1/2(x+d/2)^2/D^2)]-D[1+1/2(x-d/2)^2/D^2]`
On expanding it binomially, we get
`S_2P-S_1P=1/(2D)[4xd/2]=(xd)/D`
For constructive interference, the path difference is an integral multiple of wavelengths, that is, path difference is nλ.
`:.nlambda=(xd)/D`
`x=(nlambdaD)/d`
where n = 0, 1, 2, 3, 4, …
Similarly, for destructive interference,
`x_n=(2n-1)lambda/2D/d`
Graph of Intensity Distribution in Young’s Double-Slit Experiment
(ii) On comparing the interference pattern observed in Young's double slit experiment (interference) with single-slit diffraction pattern (diffraction), we can have three distinguishing features:
In the interference pattern, all the bright fringes have the same intensity. In a diffraction pattern, all the bright fringes are not of the same intensity
In the interference pattern, the dark fringe has zero or very small intensity so that the bright and dark fringes can easily be distinguished. In diffraction pattern, all the dark fringes are not of zero intensity
In the interference pattern, the widths of all the fringes are almost the same, whereas in diffraction pattern, the fringes are of different widths
संबंधित प्रश्न
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.
A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.
How is the fringe width of an interference pattern in Young's double-slit experiment affected if the two slits are brought closer to each other?
A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flight in two cases, then what is the product of two times of flight?
In Young's double slit experiment shown in figure S1 and S2 are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has strong intensity?
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?
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is ______.
- 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.
