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प्रश्न
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.
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उत्तर
The angular width of the central maximum obtained on the screen.
`=(2λ)/d`
`=(2×500×10^-9)/(0.2×10^-3)`
`=5×10^-3 m`
The number of fringes obtained in Young's double slit experiment
`"Angular width"/"Fringe width"=(5×10^-3)/(0.5×10^-3)=10`
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What should be the path difference between two waves reaching a point for obtaining constructive interference in Young’s Double Slit experiment ?
In Young's double-slit experiment, the two slits are separated by a distance of 1.5 mm, and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths of 650 nm and 520 nm is used to obtain interference fringes.
Find the distance of the third bright fringe for λ = 520 nm on the screen from the central maximum.
Write the conditions on path difference under which constructive interference occurs in Young’s double-slit experiment.
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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