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
Show that the angular width of the first diffraction fringe is half that of the central fringe.
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.
If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
In a Young's double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron (1 micron = 10−6 m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
A Young's double slit apparatus has slits separated by 0⋅28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by red light \[\left( \lambda = 700\text{ nm in vacuum} \right).\] Find the fringe-width of the pattern formed on the screen.
A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle \[\theta = \sin^{- 1} \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.
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 of wavelength 600 nm, 5th bright fringe is at a distance of 0·48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e. fringe separation.
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
