in young's double slits experiment while using a source of light of wavelength 4500A^0 ,the fringe width obtained is 0.4.if the distance between the slits and the screen is reduced of half calculate the new fringe width.
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Explanation:
ANSWER
β=
d
Dλ
β
1
=
d
D
1
λ
1
0.6×10
−2
=
d
D
1
5000×10
−10
⇒
D
1
d
=
0.6×10
−2
5000×10
−10
β
2
=0.003m
β
2
=
d
D
2
λ
2
0.003=
2d
D
2
λ
2
(D
2
=
2
1
D
1
as given in problem)
0.003=
2d
D
1
λ
2
λ
2
=
D
1
0.003×2×d
=0.003×2×
0.6×10
−2
5000×10
−10
(From1)
=50×10
−8
=5×10
−7
m
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