University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2021
Views:

361

Pages:

6
Author:

Dixie Summers

Newton's Rings Abstract: When parallel monochromatic light is incident normally on a plano-convex lens placed on a reflective glass plate, light will refract and reflect, creating alternating bright and dark rings. This is known as Newton's Rings. They are an example of fringes with equal thickness. We can use this to calculate the radius of curvature of the lens or to calculate the refractive index of a transparent liquid that fills the gap between the lens and glass plate. Introduction: Experiment Set-ups: experimental setup with sodium lamp, convex lens, beam splitter, air film, glass plate, and microscope To create monochromatic parallel rays, we use a sodium lamp placed at the focal point of a convex lens. The beam splitter will direct the rays normally at the plane side of the plano-convex lens. Rays will pass without being refracted first. After passing towards the convex end, rays will refract as it passes to the air film (or the transparent liquid) and gets reflected partially. The refracted light will later be reflected by the glass plate to get refracted again. Then it gets refracted as it passes the lens again. The rays then interfere and the pattern is visible on the microscope. The interface between ray 1 and 2 causes the fringes. The pattern visible under the microscope: Notice that the rings get thinner as we move away from the center. The middle ring is dark because it gets reflected on itself. Film Thickness: The condition for a bright fringe: 2t = mλ, m = 0, 1, 2, ... where: t = thickness of film λ = wavelength of light m = order of the fringe Since we use air as the refractive index of the film: 2t = (m + 1/2)λ Using the Pythagorean Theorem: R² = (R-t)² + r² where: R = radius of curvature of the lens r = radius of mth dark fringe By rearranging, we get: R = (r² + 2tm) / 2t By measuring r and m for various dark fringes, we can calculate the radius of curvature of the lens. Newton's rings are an example of fringes of equal thickness. Fringes of equal thickness are a result of light being reflected and refracted at the boundaries of the 1 film. For the case of Newton's rings, Newton's rings are useful to find the radius of curvature of a plano-convex lens and/or the refractive index of a transparent liquid. The formation of fringes as rings is due to the shape of the spherical lens. The term "equal thickness" refers to the film's thickness along one ring, the thickness is equal. Conclusion The interference between rays is constructive if the path difference = mλ. Due to reflection, there is a phase shift of λ/2. The middle fringe is dark. The fringes are a result of interference between light that gets reflected and the rays that did pass and get reflected later by the glass plate. We can find the radius of curvature using: mλR / (m + 1/2) = R We can find n using: n = (r_air / r_fluid)^2 The condition for: constructive (bright): rm^2 = (m + 1/2)λR destructive (dark): rm^2 = mλR Along one ring, the thickness of our film is equal.