The light of wavelength 6,000 Å falls normally on a slit of width 2 mm. Calculate the linear width of central maximum on a screen of 4 m away from the slit.
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Fringe width is given by β=
d
λD
In the first case, λ
1
=6000 A
∘
=6000×10
−10
m
β
1
=6 mm=0.006 m
∴0.006=6000×10
−10
d
D
....(1)
In the second case, the distance between the slits and the screen is
2
D
,β
2
=4 mm=0.004 m
∴0.004=λ
2
×
d
2
D
....(2)
Directing equations (2) by (1) we get
0.006
0.004
=
2×6000×10
−10
λ
2
λ
2
=
0.006
0.004×2×6000×10
−10
λ
2
=
0.006
48×10
−10
λ
2
=8000×10
−10
λ
2
=8000 A
∘
.
I hope this helps you
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