Answer to Question #98046 in Optics for Akanksha dubey

Question #98046
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.
1
Expert's answer
2019-11-07T09:10:55-0500

We can write the condition for minimum interference as

2hn+λ2=(2m+1)×λ2(1)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)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=hl(3)θ=\frac {h}{l} (3)


Using (3) we got

h=θ×l(4)h= θ×l (4)

We put (4) in (2)

2θln=mλ(5)2θln=mλ (5)

We got

ml=2θnλ(6)\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

ml=3.68fringescm\frac {m}{l}=3.68 \frac {fringes}{cm}



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