Osm Physics by Ashish

Fizeau’s Fringes Fizeau’s fringes are fringes of equal thickness produced by the interference of light reflected from the two surfaces of a thin wedge-shaped film . They are generally straight, parallel, and equally spaced . They are localized near the film. The thickness of the film varies from point to point. The optical path difference depends on the thickness of the film . For a film of refractive index μ , thickness t , and refracted angle r : Δ = 2μt cos r For a wedge-shaped air film at nearly normal incidence: Δ = 2t The condition for dark fringes in reflected light is: 2t = mλ If the wedge angle is θ , then: t = xθ Therefore: 2xθ = mλ The fringe width is: β = λ / (2θ) Difference Between Haidinger’s and Fizeau’s Fringes Property Haid...

1. Determination of Wave Length of a Monochromatic Light Source In Newton’s experiment if we use a light source of unknown wave length (say sodium lamp) then we can determine the wavelength of light source by measuring the diameters of Newton’s ring. If D n is diameter of nth dark ring formed due to air film then D n 2 = 4nλR Where n is any integer number. Similarly if D n+p is the diameter of ( n+p ) th ring D n+p 2 = 4(n + p)λR Using this equation, we can write D n+p 2 − D n 2 = 4(n + p)λR − 4nλR = 4pλR or λ = D n+p 2 − D n 2 ⁄ 4pR      ...... (5.20) Where p is any integer number and R is radius of curvature of plano-convex lens. 2. Determination of Refractive Index of a Liquid by Newton’s Rings Experiment In Newton’s rings experiment the diameter of n th dark ring...

Newton's rings in a special case of wedge shaped film in which an air film is formed between a glass plate and a convex surface of lens. The thickness of air film is zero at the center and increases gradually towards the outside. When a plano-convex lens of large focal length is placed on a plane glass plate, a thin air film is formed between the lower surface of plano-convex lens and upper surface of glass plate. When a monochromatic light falls on this film the light reflected from upper and lower surfaces of air film, and after interference of these rays, we get an inner dark spot surrounded by alternate bright and dark rings called Newton's rings. These rings are first observed by Newton and hence called Newton's rings. Experimental Arrangement for Reflected Light The experimental arrangement for Newton's rings experiment. A beam of monochromatic source S is made parallel by using a convex lens L. The parallel ...

The two surfaces of the thin film are slightly inclined to each other at an angle \( \theta \). Therefore, the thickness of the film gradually increases from one end to the other. The ray \(SA\) is the incident ray falling on the upper surface of the thin film at \(A\). At \(A\), the incident light is partially reflected and partially transmitted. A part of the incident light is reflected directly from the upper surface at \(A\) and travels along \(AR_1\). This is the first reflected ray. The remaining part of the incident ray enters the thin film at \(A\). It travels inside the film and reaches the lower surface at \(B\). At \(B\), a part of the light is reflected back towards the upper surface. It reaches the upper surface at \(C\) and emerges along \(CR_2\)...

Haidinger's Fringes Haidinger's fringes are a system of concentric circular interference fringes produced by the interference of two beams whose path difference depends on their angle of inclination . They are also known as fringes of equal inclination . Formation in a Michelson Interferometer In a Michelson interferometer, let \(d\) be the separation between the two reflecting surfaces and let a ray make an angle \(\theta\) with the normal. The optical path difference between the two interfering beams is: \[ \Delta = 2d\cos\theta \] Condition for Bright Fringes For constructive interference, the condition for bright fringes is: \[ 2d\cos\theta = n\lambda \] where: \(d\) = separation between th...

Determination of Refractive Index or Thickness A thin transparent film of thickness \(t\) and refractive index \(\mu\) is placed normally to the light ray between the mirror \(M_2\) and the beam splitter plate. Due to the insertion of the film, the optical path of the light ray increases by \[ 2(\mu - 1)t \] Consequently, the white fringe is displaced from its original position. The displacement of mirror \(M_1\) is adjusted with the help of a micrometer screw until the cross-wire of the telescope again coincides with the white fringe. Let the displacement of mirror \(M_1\) be \(x\). The optical path difference produced due to the insertion of the thin film becomes equal to the optical path difference produced by the displacement of mirror \(M_1\). Since ...

Application of Michelson’s Interferometer Michelson’s interferometer can be used to determine the difference in wavelengths (or frequencies) between two closely spaced spectral lines . This method is based on the periodic change in visibility of interference fringes produced by the two wavelengths. Let the wavelengths of two spectral lines be: \[ \lambda_1 \quad \text{and} \quad \lambda_2 \] where \( \lambda_1 \) and \( \lambda_2 \) are very close to each other. Principle When light of two nearly equal wavelengths is used in a Michelson interferometer, two sets of circular interference fringes are formed. Due to the slight difference in wavelengths, the bright fringes of one system gradually coincide with the dark fringes of the other system. Consequently, the fringes alter...

Measurement of Wavelength Using Michelson Interferometer Application of Michelson's Interferometer Measurement of Wavelength One of the important applications of a Michelson interferometer is the accurate measurement of the wavelength of monochromatic light. For a dark finge One of the important applications of a Michelson interferometer is the accurate measurement of the wavelength of monochromatic light. \[ 2d = n\lambda \\.............(1)\] When the movable mirror \(M_1\) is displaced through a distance \(x\), the optical path difference changes by \(2d\), because the light travels to the mirror and returns. As a result, the interference fringes shift. If \(N\) fringes cross the reference point during the displacement of the mirror, then the change in optical path is: \[ ...