In a Young\'s double-slit experiment, the wavelength of the lightused is 500 nm
ID: 1725728 • Letter: I
Question
In a Young's double-slit experiment, the wavelength of the lightused is 500 nm (in vacuum), and theseparation between the slits is 3.80 x10-6 m. Determine the angle that locateseach of the following. (a) the dark fringe for which m =0°
(b) the bright fringe for which m = 1
°
) the dark fringe for which m = 1
°
(d) the bright fringe for which m = 2
° (a) the dark fringe for which m =0
°
(b) the bright fringe for which m = 1
°
) the dark fringe for which m = 1
°
(d) the bright fringe for which m = 2
°
Explanation / Answer
The condition for the bright fringe is dsin =m and for dark fringe dsin = (m+1/2)
(a) for first dark fringe m = 0 therefore dsin = (1/2) so = sin-1(/2d) =sin-1(500*10-9m/2*3.8*10-6m) =3.77o
(b) for first bright fringe m = 1 dsin = so = sin-1(/d) =sin-1(500*10-9m/3.8*10-6m) =7.56o
(c) for dark fringe m = 1 therefore dsin = (3/2) so = sin-1(3/2d) =sin-1(3*500*10-9m/2*3.8*10-6m) =11.38o
d) for bright fringe m = 2 therefore dsin = 2 so = sin-1(2/d) =sin-1(2*500*10-9m/3.8*10-6m) =15.26
Hope this helps you.
(b) for first bright fringe m = 1 dsin = so = sin-1(/d) =sin-1(500*10-9m/3.8*10-6m) =7.56o
(c) for dark fringe m = 1 therefore dsin = (3/2) so = sin-1(3/2d) =sin-1(3*500*10-9m/2*3.8*10-6m) =11.38o
d) for bright fringe m = 2 therefore dsin = 2 so = sin-1(2/d) =sin-1(2*500*10-9m/3.8*10-6m) =15.26
Hope this helps you.
(c) for dark fringe m = 1 therefore dsin = (3/2) so = sin-1(3/2d) =sin-1(3*500*10-9m/2*3.8*10-6m) =11.38o
d) for bright fringe m = 2 therefore dsin = 2 so = sin-1(2/d) =sin-1(2*500*10-9m/3.8*10-6m) =15.26
Hope this helps you.
d) for bright fringe m = 2 therefore dsin = 2 so = sin-1(2/d) =sin-1(2*500*10-9m/3.8*10-6m) =15.26
Hope this helps you.