Question

A beam of Monochromatic light of wavelength 582 nm falls normally on a glass wedge with the wedge angle 20'' of arc. If the refractive index of the glass is 1.5, find the number of dark fringes per cm of the wedge length.

Answer 1

The condition for interference is

$${2}{t}={m}{\lambda} (1)$$

where t is the thickness of the air wedge at a distance x from the point of contact of the glass plates.

θ is the angle between the plates in radians (this angle is small, so tan θ = θ in radians). In this case, we got

$$\theta=\frac{t}{x} (2)$$

So

$$2{\theta}{x}={m}{\lambda} (3)$$

We got

$$\frac{m}{x}=\frac{2\theta}{\lambda} (4)$$

20'' is equal to 9.59×10^-5 radians

Using (4) we get 3.29 fringes/cm

Answer:

3.29 fringes/cm