Question

In a double slit experiment, the slit
separation is 0.20 cm and the slit to screen
distance is 100 cm. The positions of the
first three minima, if wavelength of the
source is 500 nm is​

Answer 1

ANSWER

dy/D=nλ

I

st

minima=

2×0.2×10

−2

1×500×10

−9

=1250×10

−7

=±0.125×10

−3

m

=0.0125cm

II

nd

minima=3×I

st

minima

=3×0.0125cm

=±0.0375cm

III

rd

minima=5×I

st

minima

=5×0.0125cm

=±0.0625cm.