Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: Medium

State giving the reasons what change do you expect in newton’s ring

(i) if top surface of the glass plate on which the lens is kept is highly silvered.

(ii) if the lens is lifted slowly off the plate by a distance of half the wavelength.

(i) If the top surface of glass plate on which the lens is kept is highly silvered or replaced by plane mirror, the fringes will disappear and we shall get a uniform illumination. It is because the intensity of light reflected from the surface of the mirror is very large as compared to the intensity of light reflected from the curved surface of air film.

  (ii) If the lens is lifted slowly off the plate by a distance of half the wavelength, then nth dark ring will shrink to (n-1)th  dark ring. Thus, as the distance between the lens and plate is increased the thickness and hence the order of ring at a given point increases. The rings therefore come closer until they are no longer separately observed.

We know radius of nth ring is $r_n^ 2$ = 2Rt

 Therefore 2t = $r_n^ 2$  / R--------------------------------(1)

For the condition of darkness, 2μtcos(r+ϴ ) =nλ

For normal incidence and small angle cos(r+ϴ ) =1 and for air film μ =1

Therefore 2t =nλ-------------------------------------(2)

Now if the lens is raised by some distance say y then t = (t+y) where y =λ/2

Substituting value of t in equation (2)

i.e 2(t+y) = nλ

2t +2y = nλ

Substituting value of 2t From equation (1)

$r_n^2$ / R = nλ -2y

$r_n$ = $\sqrt{(R(nλ-2y) )}$

the above equation represent radius of nth ring when lens is lifted up through a distance of ‘y’

let y = λ/2

substituting y as half the wavelength and for n =1 , the radius of first dark ring becomes zero. That means the first dark ring merges into centre.

$r_1$ = $\sqrt{(R(1λ-2λ/2) )}$

Therefore $r_1$ = 0.

If n is substituted as n=2, then second dark ring occupy the position of s first dark ring and so on. Thus nth dark ring will shrink to (n-1)th dark ring.