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प्रश्न
If Young's double slit experiment is performed in water, _________________ .
पर्याय
the fringe width will decrease
the fringe width will increase
the fringe width will remain unchanged
there will be no fringe
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उत्तर
the fringe width will decrease
As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,
Here,
\[\lambda_M = \lambda/\eta\]
\[ \lambda_M = \text{wavelength in medium}\]
\[\lambda = \text{wavelength in vacuum}\]
\[\eta = \text{refractive index of medium}\]
Hence, fringe width decreases when Young's double slit experiment is performed under water.
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संबंधित प्रश्न
In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.
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Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2.
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is a distance of 2.5 mm away from the centre. Find the width of the slit.
Write two characteristics features distinguish the diffractions pattern from the interference fringes obtained in Young’s double slit experiment.
What is the effect on the interference fringes to a Young’s double slit experiment when
(i) the separation between the two slits is decreased?
(ii) the width of a source slit is increased?
(iii) the monochromatic source is replaced by a source of white light?
Justify your answer in each case.
In Young’s experiment interference bands were produced on a screen placed at 150 cm from two slits, 0.15 mm apart and illuminated by the light of wavelength 6500 Å. Calculate the fringe width.
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.
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.
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).
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ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
REASON (R): Fringe width is proportional to (d/D).
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