Question

Two narrow slits separated by 0.30 mm are illuminated with light of wavelength 496 nm. (a) How far are the first three bright fringes from the center of the pattern if observed on the screen 130 cm distant? (b) How far are the first three dark fringes from the center of the pattern?

Answer 1

Answer:

Explanation:

a)

d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m

λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m

n = order of the bright fringe

D = screen distance = 130 cm = 1.30 m

$x_{n}$ = Position of nth bright fringe

Position of nth bright fringe is given as

$x_{n} =\frac{ n D \lambda }{d}$

For n = 1

$x_{1} =\frac{ (1) (1.30)(496\times 10^{-9})}{0.30\times 10^{-3}}$

$x_{1} = 2.15\times 10^{-3}m$

For n = 2

$x_{2} =\frac{ (2) (1.30)(496\times 10^{-9})}{0.30\times 10^{-3}}$

$x_{2} = 4.30\times 10^{-3}m$

For n = 3

$x_{2} =\frac{ (2) (1.30)(496\times 10^{-9})}{0.30\times 10^{-3}}$

$x_{2} = 6.45\times 10^{-3}m$

b)

Position of nth dark fringe is given as

$y_{n} =\frac{ (2n+1) D \lambda }{2d}$

For n = 1

$y_{1} =\frac{ (2(1)+1) (1.30)(496\times 10^{-9})}{2(0.30\times 10^{-3})}$

$y_{1} = 3.22\times 10^{-3}m$

For n = 2

$y_{2} =\frac{ (2(2)+1) (1.30)(496\times 10^{-9})}{2(0.30\times 10^{-3})}$

$y_{2} = 5.4\times 10^{-3}m$

For n = 3

$y_{3} =\frac{ (2(3)+1) (1.30)(496\times 10^{-9})}{2(0.30\times 10^{-3})}$

$x_{3} = 7.5\times 10^{-3}m$