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In a Young’s double slit experiment, the distance between the two slits is 0.5 mm and the distance between the screen and the slits is 1 m. When a light of wavelength 500 nm is incident on the slits, what would be distance between the two second bright fringes?

In a Young’s double slit experiment, the distance between the two slits is 0.5 mm and the distance between the screen and the slits is 1 m. When a light of wavelength 500 nm is incident on the slits, what would be distance between the two second bright fringes? Correct Answer 2 mm

As we know, β = λD/d. In this question, we need to find 2 X β. Here, d = 0.5 mm and D = 1 m Therefore, β = 500 nm / 0.5 mm = 1 mm Now, Separation between the second bright fringe on both sides of the central maxima = 2 X 1 mm = 2 mm.

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