Question

3. In a Young’s double-slit experiment the distance between the slits is 1 mm and the distance of the screen from the slits is 1m. If light of wavelength 6000 A0
is used, find the distance between the second dark fringe and the fourth bright fringe.

Answer 1

Explanation:

given

d=1mm

D=1m

λ=6000Ao

second dark fringe=

y=(2n-1)/2×λD/d

n=2

y=3/2×6000×1/1×10^-3

y1=9m

fourth bright fringe

y=nλD/d

n=4

y=4×6000×1/1×10^-3

y2=24

distance b/w them is y2-y1

=24-9

=15m