Question

Light from a coherent monochromatic light source with a wavelength of 5.20 ✕ 102 nm is incident on (and perpendicular to) a pair of slits separated by 0.310 mm. An interference pattern is formed on a screen 2.10 m from the slits. Find the distance (in mm) between the first and second dark fringes of the interference pattern.

Answer 1

Answer:

y = 3.52 mm

Explanation:

given,

wavelength of the light source (λ)= 520 nm

slits is separated (d) = 0.310 mm

distance to form interference pattern(D) = 2.10

distance between first and second dark fringe = ?

now, using displacement formula

 $y = \dfrac{m\lambda\ D}{d}$

where y is the fringe width

for first and second dark fringe

now, m = 1

 $y = \dfrac{1\times \lambda\ D}{d}$

 $y = \dfrac{1\times 520 \times 10^{-6}\ 2100}{0.310}$

     y = 3.52 mm

width of the first bright fringe is equal to 3.52 mm

hence, distance between first and second dark fringe is equal to                   y = 3.52 mm