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Q. In Young's double slit experiment carried out with
light of wavelength $ (\lambda ) = 5000 \mathring{A}$ , the distance
between the slits is 0.2 mm and the screen is at 200 cm
from the slits. The central maximum is at .v = 0. The
third maximum (taking the central maximum as
zeroth maximum) will be at x equal to

AIPMTAIPMT 1992

Solution:

$ x= (n) \lambda \frac{D}{d} = 3\times 5000 \times {10}^{-10} \times \frac{2}{0.2\times {10}^{-3}} $
$= 1.5 \times {10}^{-2} m = 1.5 cm$