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प्रश्न
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 ______.
विकल्प
there shall be alternate interference patterns of red and blue.
there shall be an interference pattern for red distinct from that for blue.
there shall be no interference fringes.
there shall be an interference pattern for red mixing with one for blue.
उत्तर
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 there shall be no interference fringes.
Explanation:
Condition for Observing Interference
The initial phase difference between the interfering waves must remain constant. Otherwise, the interference will not be sustained.
The frequency and wavelengths of two waves should be equal. If not the phase difference will not remain constant and so the interference will not be sustained.
The light must be monochromatic. This eliminates overlapping of patterns as each wavelength corresponds to one interference pattern.
Here in this problem of Young’s double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case, due to filtration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.
The wavefront emitted by a narrow source is divided in two parts reflection, refraction or diffraction. The coherent sources so obtained are imaginary.
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In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width.
Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing λ = 400 nm). Describe the nature of the fringe pattern observed.
The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be ____________ .
Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to
(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
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A beam of light consisting of two wavelengths 600 nm and 500 nm is used in Young's double slit experiment. The silt separation is 1.0 mm and the screen is kept 0.60 m away from the plane of the slits. Calculate:
- the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
- the least distance from the central maximum where the bright fringes due to both wavelengths coincide.
- 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 an interference experiment, a third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point?
