A helium-neon laser produces light with a wavelength of 638 nm. when this light is shone though a double slit apparatus, an interference pattern is produced on a screen 2.0 m away, with the distance between the first and seventh nodal lines being 5.0 cm.
(a) determine the distance between the slits.
(b) what is the maximum number of bright fringes that could possibly appear on the screen?
(a) We can use the double-slit interference formula:
d sin(theta) = m(lambda)
where d is the distance between the two slits, theta is the angle between the central bright fringe and the mth bright fringe, and m is the order of the bright fringe.
Since we know the distance between the first and seventh nodal lines, we can count the number of bright fringes between them (which is 6). The distance between adjacent bright fringes is given by
delta y = lambda*D/d
where D is the distance from the slits to the screen. We know lambda and D, so we can solve for d:
d = lambda*D/delta y = (638 nm)*(2.0 m)/(0.05 m) = 25.5 micrometers
(b) The maximum number of bright fringes that can appear on the screen is given by
N = D/delta y
where delta y is the distance between adjacent bright fringes, as above. Substituting, we find
N = (2.0 m)/(638 nm/0.05 m) = 157
So the maximum number of bright fringes is 157. Note that this assumes there is no diffraction limit on the number of fringes that can be resolved. In reality, there may be limitations due to the finite size of the slits and the wavelength of light used.