optics
Derive Newton ring experiment and find its diameter no spam ans i want
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Answers
Answer:
{\displaystyle t={r^{2} \over 2R}}
Explanation:
......
Answer:
When a parallel beam of monochromatic light is incident normally on a combination of a
plano-convex lens L and a glass plate G, as shown in Fig.1, a part of each incident ray is
reflected from the lower surface of the lens, and a part, after refraction through the air film between
the lens and the plate, is reflected back from the plate surface. These two reflected rays are coherent,
Fig. 2. Newton’s rings
Date :
32
Newton’s Rings
hence they will interfere and produce a system of alternate dark and bright rings with the point of
contact between the lens and the plate as the center. These rings are known as Newton’s ring.
For a normal incidence of monochromatic light, the path difference between the reflected
rays (see Fig.1) is very nearly equal to 2t where and t are the refractive index and thickness of the
air-film respectively. The fact that the wave is reflected from air to glass surface introduces a phase
shift of . Therefore, for bright fringe
2
1 2 t n
; n = 0,1,2,3 (1)
and for dark fringe
2t n
; n = 0,1,2,3 (2)
For n-th (bright or dark) ring (see Fig. 2), we also have
2 2
2
4
R t R
D n
(3)
where Dn = the diameter of the n-th ring and R = the radius
of curvature of the lower surface of the plano-convex lens.
On neglecting t
2
, equation (3) reduces to
Dn
2
= 8tR (4)
From equations (1) and (4), we get,
μ
λ R
2
1 D 4 n
2
n
, for n-th bright ring (5)
μ
λ R
2
1 D n m 4 n m
2
,
for (n+m)-th bright ring (6)
Similarly, from equations (2) and (4), we obtain
n R
Dn
2 4
, for n-th dark ring (7)
n m R
Dn m
4( )
, for (n+m)-th dark ring (8)
Thus for bright as well as dark rings, we obtain
4mλ
μ D D
R
2
n m n
2
… (9)
Since =1 for air-film, above equation gives
4mλ