Question

In young's double slit experiment of slits 0.18mm apart are illuminated by light of wavelength 589.3nm.calculate the distance of I) 5th bright fringe and ii) 3rd dark fringe from the midpoint of the interference pattern obtained on the screen kept at a distance of 0.6m from slits.

Answer 1

i) 5th bright fringe 9.8 mm

ii)3rd dark fringe 4.9 mm

Answer 2

Answer: The distance of the 3rd dark fringe from the midpoint of the interference pattern is 0.0056117 m.

In Young's double-slit experiment, the bright fringes are formed when the path difference between the two slits is an integer multiple of the wavelength, while the dark fringes are formed when the path difference is a half-integer multiple of the wavelength.

Let the distance between the two slits be d = 0.18 mm = 0.00018 m, the wavelength of the light be λ = 589.3 nm = 0.0005893 m, and the distance between the slits and the screen be L = 0.6 m.

The distance between the central maximum and the nth bright fringe is given by:

                          y_bright = (n * λ * L) / d

Substituting the given values, we get:

y_5th bright = (5 * 0.0005893 * 0.6) / 0.00018 = 0.009837 m

So, the distance of the 5th bright fringe from the midpoint of the interference pattern is 0.009837 m.

The distance between the central maximum and the nth dark fringe is given by:

                             y_dark = ((2 * n - 1) * λ * L) / (2 * d)

Substituting the given values, we get:

y_3rd dark = ((2 * 3 - 1) * 0.0005893 * 0.6) / (2 * 0.00018) = 0.0056117 m

So, the distance of the 3rd dark fringe from the midpoint of the interference pattern is 0.0056117 m.

Learn more about young's double-slit experiment here

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