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प्रश्न
The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.
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उत्तर
Let w, a,I represent the slit width, amplitude and intensity.
`I_(min)/I_(max)=(a_1-a_2)^2/(a_1+a_2)^2=9/25`
`((a_1-a_2))/((a_1+a_2))=3/5`
Or
`a_1/a_2=4/1`
and
`w_1/w_2=(a_2)^2/(a_2)^2=16/1`
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संबंधित प्रश्न
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REASON (R): Fringe width is proportional to (d/D).
In a double-slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 × 10-2 m towards the slits, the change in fringe width is 3 × 10-3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ______ nm.
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- 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).
