Young’s double-slit experiment – problems and solutions

1. d is the distance between 2 slits, L is the distance between the slit and the viewing screen, P2 is the distance between the second-order fringe and the center of the screen. Determine the wavelength of light (1 Å = 10-10 m).

Known :Young’s double-slit experiment - problems and solutions 1

Distance between two slits (d) = 1 mm = 1 x 10-3 m

Distance between slit and the viewing screen (L) = 1 m

Distance between the second-order fringe and the central fringe (P2) = 1 mm = 1 x 10-3 m

Order (n) = 2

Wanted : the wavelength of light (λ)

Solution :

The equation of double-slit interference (constructive interference) :

d sin θ = n λ

sin θ ≈ tan θ = P2 / L = (1 x 10-3) / 1 = 1 x 10-3 m

The wavelength of light :

λ = d sin θ / n

λ = (1 x 10-3)(1 x 10-3) / 2 = (1 x 10-6) / 2

λ = 0.5 x 10-6 m = 5 x 10-7 m

λ = 5000 x 10-10 m

λ = 5000 Å

Read :  Determine vector components

2. Figure below shown result of a double-slit interference. Determine the wavelength of light (1 m = 1010 Å)

Known :

Distance between two slits (d) = 0.8 mm = 8 x 10-4 mYoung’s double-slit experiment - problems and solutions 2

Distance between slit and the viewing screen (L) = 1 m

Distance between the fourth-order fringe and the central fringe (P) = 3 mm = 3 x 10-3 m

Order (n) = 4

Wanted : The wavelength of light (λ)

Solution :

The equation of double-slit interference (constructive interference) :

d sin θ = n λ

sin θ ≈ tan θ = P / L = (3 x 10-3) / 1 = 3 x 10-3 me

The wavelength of light :

λ = d sin θ / n

λ = (8 x 10-4)(3 x 10-3) / 4 = (24 x 10-7) / 4

λ = 6 x 10-7 m = 6000 x 10-10 m

λ = 6000 Å

Read :  Conservation of mechanical energy – problems and solutions

3. Based on figure below, point A and B is the first two bright interference fringes and the wavelength of light is 6000 Å (1 Å = 10-10 m). Determine distance between two slits.

Known :

Distance between slit and the viewing screen (L) = 1 mYoung’s double-slit experiment - problems and solutions 3

The wavelength of light (λ) = 6000 Å = 6000 x 10-10 m = 6 x 10-7 m

Distance between the first-order fringe and the central fringe (P) = 0.2 mm = 0.2 x 10-3 m = 2 x 10-4 m

Order (n) = 1

Wanted : Distance between two slits (d)

Solution :

The equation of constructive interference :

d = n λ / sin θ

sin θ ≈ tan θ = P / L = (2 x 10-4) / 1 = 2 x 10-4 m

Distance between two slits :

d = n λ / sin θ = (1)(6 x 10-7) / (2 x 10-4)

d = (6 x 10-7) / (2 x 10-4) = (3 x 10-3)

d = 0.003 m

d = 3 mm

Read :  Motion on the rough inclined plane with the friction force - application of Newton's law of motion problems and solutions