Advertisements
Advertisements
प्रश्न
A double slit S1 − S2 is illuminated by a coherent light of wavelength \[\lambda.\] The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (see the following figure). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.
Advertisements
उत्तर
Given:-
Separation between the two slits = d
Wavelength of the coherent light =λ
Distance between the slit and mirror is D1.
Distance between the slit and screen is D2.
Therefore,
apparent distance of the screen from the slits,
\[D = 2 D_1 + D_2 \]
Fringe width, \[\beta = \frac{\lambda D}{d} = \frac{\left( 2 D_1 + D_2 \right) \lambda}{d}\]
Hence, the required fringe width is \[\frac{\left( 2 D_1 + D_2 \right) \lambda}{d}.\]
APPEARS IN
संबंधित प्रश्न
In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is `λ/3`.
Using analytical method for interference bands, obtain an expression for path difference between two light waves.
In Young’s experiment, the ratio of intensity at the maxima and minima in an interference
pattern is 36 : 9. What will be the ratio of the intensities of two interfering waves?
A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.
Estimate the number of fringes obtained in Young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit.
The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
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.
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?

Draw a neat labelled diagram of Young’s Double Slit experiment. Show that `beta = (lambdaD)/d` , where the terms have their usual meanings (either for bright or dark fringe).
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?
In Young’s double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other?
Two slits in Young's interference experiment have width in the ratio 1 : 2. The ratio of intensity at the maxima and minima in their interference is ______.
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 ______.
Why is the diffraction of sound waves more evident in daily experience than that of light wave?
Using Young’s double slit experiment, a monochromatic light of wavelength 5000Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000Å is used and the separation between the slits is doubled, then the new fringe width will be ______.
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 :
In Young's double-slit experiment, the screen is moved away from the plane of the slits. What will be its effect on the following?
- The angular separation of the fringes.
- Fringe-width.
