Answer to Question #98046 in Optics for Akanksha dubey

Question

A glass-wedge with the angle of wedge of 30 seconds of an arc is formed With a liquid of

R.1. 1.47. Find the number of dark interterence fringes per cm of the wedge length.

Answer 1

We can write the condition for minimum interference as

$2hn+\frac {λ}{2}=(2m+1)×\frac {λ}{2} (1)$

where h is the thickness of the wedge, m is the number of dark interterence fringes

Using (1) we got

$2hn=mλ (2)$

θ is the angle between the plates in radians (this angle is small, so \tan {θ} = θ in radians).

In this case, we got

$θ=\frac {h}{l} (3)$

Using (3) we got

$h= θ×l (4)$

We put (4) in (2)

$2θln=mλ (5)$

We got

$\frac {m}{l}=\frac {2θn}{λ} (6)$

In our case, we have θ=1.46×10^-4, n=1.47, λ=582 nm

The number of dark interterence fringes per cm of the wedge length is equal to

$\frac {m}{l}=3.68 \frac {fringes}{cm}$