Question

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Ans. 47
19. In an experiment with sodium light (5890 A),
interference pattern is obtained in which 20
spaced fringes occupy 2.30 cm on the screen.
replacing sodium light with another monochromat.
source, 30 fringes occupy 2.80 cm on the screen. Fini
the wavelength of the light from the second source​

Answer 1

Answer:

this is yr answer

Explanation:

d is the distance between the slits. ) has a direct dependence on D, fringe width decreases if the distance between the slits and screen is decreased. ) of the source is increased because, fringe width and frequency have inversely dependent on each other.If the fringe width is 0.75 mm, calculate the wavelength of light. Given: Distance between slits = d = 0.8 mm = 0.8 x 10-3 m = 8 x 10-4 m. Distance between slit and screen = D = 1.2 m, Fringe width = X = 0.75 mm = 0.75 x 10-3 m = 7.5 x 10-4 m. To Find: Wavelength of light used = λ =?∴ β decreases i.e., fringe width decreases. ( c) When separation between slits is increased, d increases. Therefore fringe width β decreases. As source slit is brought closer to double slit plane, S decreases, teh interference pattern gets less and less sharp.