Question

In a Newton’s ring experiment, the diameter of the 20th dark ring was found to be 5.82 mm and the 10th ring is 3.36mm. If the radius of the plano convex lens is 1m, calculate the wavelength of light used.

Answer 1

Hint: Circular interference fringes produced by enclosing a thin air film of varying thickness between the surface of a convex lens of a large radius of curvature and a plane glass plate are known as Newton’s rings. The monochromatic light which produces these rings is given by the following mentioned formula.

Formula Used:
The wavelength of light is given by
$\lambda = \dfrac{{r_{n + m}^2 - r_n^2}}{{mR}}$
where R is the radius of curvature of the surfaces of the lens in contact with the glass plate.
rn and rn+m are the radiuses of the nth and (n+m)th dark fringe.

Complete step by step answer:
Circular interference fringes produced by enclosing a very thin air film of varying thickness between the surface of a convex lens of a large radius of curvature and a plane glass plate are known as Newton’s rings.
The radius of dark fringes:
For a minimum intensity or a dark fringe, the mathematical condition is:
\eqalign{ & 2\mu t = n\lambda \cr & \therefore 2\mu \dfrac{{{r^2}}}{{2R}} = n\lambda \cr & {\text{or }}r_n^2 = \dfrac{{n\lambda R}}{\mu } \cr}
Given that:
Diameter of the 20th dark ring, ${D_{20}} = 5.8mm = 5.82 \times {10^{ - 3}}m$
Radius of the 20th dark ring, ${r_{20}} = 2.91 \times {10^{ - 3}}m$
Diameter of the 10th dark ring, ${D_{10}} = 3.6mm = 3.6 \times {10^{ - 3}}m$
Radius of the 10th dark ring, ${r_{10}} = 1.68 \times {10^{ - 3}}m$
Radius of plano convex lens, R=1m
We know that the wavelength of light is given by:
$\lambda = \dfrac{{r_{n + m}^2 - r_n^2}}{{mR}}$
Substituting values in the above equation we get:
\eqalign{ & \dfrac{{r_{n + m}^2 - r_n^2}}{{mR}} = \dfrac{{\left[ {{{\left( {2.91 \times {{10}^{ - 3}}} \right)}^2} - {{\left( {1.68 \times {{10}^{ - 3}}} \right)}^2}} \right]}}{{10 \times 1}} \cr & \lambda = \dfrac{{\left[ {8.4681 - 2.8224} \right] \times {{10}^{ - 6}}}}{{10}} = 0.5646 \times {10^{ - 6}} \cr & \lambda = 5646\mathop {\text{A}}\limits^ \circ \cr}
Therefore, the wavelength of light used is $5646\mathop {\text{A}}\limits^ \circ$.

Note: When a mirror is used in place of a glass plate below the lens in the Newton’s Ring experiment, no interference is observed because reflection is complete. The reflected and transmitted system of rings gets superimposed, the two systems being in complementary give rise to uniform transmission.