Question

Two extremely flat glass plates, 10.0 cm wide, touch at one end but are separated by a thin wire at the other end, forming a wedge. Light with a wavelength of 450 nm shines almost perpendicularly on the glass and forms fringes which are 1.80 mm apart. What is the diameter of the wire

Answer 1

Answer: 12.50 μm

Explanation:

Usually, reflections from the top of the bottom plate have a ½λ phase change. But they do not possess it from the bottom of the top plate (striking air). So then,

t = ½ λ = ½ x 450*10^-9m = 225*10^-9 m

So that the Wedge angle will then be

sinθ = t/1.80*10^-3 m

sinθ = 225*10^-9 / 1.80*10^-3

sinθ = 1.25*10^-4

θ = 2.18*10^-4

For the whole wedge, the hypotenuse = 0.10m, i.e the vertical wire diameter = d

Sinθ = d / 0.10

1.25^-4 = d / 0.10

d = 0.10 x 1.25*10^-4

d = 1.25*10^-5m

d = 12.50^-6 m (12.50 μm)