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Q. In Youngs double slit experiment, the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1\, m$ from the place containing the slits. If wavelength of the incident light is $6000\,\mathring{A}$ then the separation between the third bright fringe and central maxima is :

Haryana PMTHaryana PMT 2002

Solution:

Separation between, third-bright fringe and central maxima
$=\frac{3 D \lambda}{d}=\frac{3 \times 1 \times 6 \times 10^{-7}}{0.5 \times 10^{-3}}$
$=3.6 \times 10^{-3} m =3.6\, mm$