Question

young's double slit experiment while using a source of light of wavelength 5000 am strong the fringe width is 0.6 CM if the distance between the screen and the slit is reduced by the half of what should be the wavelength of the height to get the fringe width of 0.4cm​

Answer 1

Explanation:

Initial fringe width β

i

=0.6 cm=0.006 m

Final fringe width β

f

=0.003=

2

β

i

Fringe width β=

d

λD

⟹

β

i

β

f

=

λ

i

D

i

λ

f

D

f

Or

2

1

=

5000× D

i

λ

f

(D

i

/2)

⟹ λ

f

=5000 A

o