Question

If 710-nm and 655-nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.1 m away?

Answer 1

Answer:

Explanation:

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.1 m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2403.07 x 10⁻⁶ m

= 2.403 x 10⁻³ m

= 2.403 mm .

For λ = 655 nm

position = 2 λ D / d

λ = 655 nm , D = 1.1 m

d = .65 x 10⁻³

position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³

= 2216.91 x 10⁻⁶ m

= 2.217 x 10⁻³ m

= 2.217 mm .

Difference between their position

= 2.403 - 2.217 = .186 mm .