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Question
Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit.
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Solution
Intensity distribution as function of phase angle, when diffraction of light takes place through coherently illuminated single slit.
The Intensity pattern on the screen is shown in the given figure.
Width of central maximum = `(2"D"λ)/"a"`
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RELATED QUESTIONS
(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.
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.
Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits.
- Find the minimum value of d so that there is a dark fringe at O.
- Suppose d has this value. Find the distance x at which the next bright fringe is formed.
- Find the fringe-width.
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.
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.
In Young's double slit experiment the slits are 0.589 mm apart and the interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 7.5 mm from the dark fringe which is second from the center of the fringe pattern. Find the wavelength of the light used.
A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit. Find:
- The distance of the second bright fringe from the central maximum pertaining to the light of 600 nm.
- The least distance from the central maximum at which bright fringes due to both wavelengths coincide.
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, show that:
`β = (λ"D")/"d"`
Where the terms have their usual meaning.
What does \(d\) denote in Young's double-slit experiment?
