Osm Physics by Ashish

Fresnel Diffraction Definition Fresnel diffraction is the diffraction of light in which the source and screen are at finite distances from the diffracting obstacle or aperture. Theory Fresnel diffraction is based on the Huygens–Fresnel principle . According to this principle, every point on a wavefront acts as a source of secondary wavelets. These wavelets spread into the region behind the obstacle or aperture and interfere with one another. In Fresnel diffraction, the incident wavefront is generally spherical or cylindrical . Therefore, the curvature of the wavefront cannot be neglected. Fresnel Half-Period Zones The wavefront is divided into a number of zones called Fresnel half-period zones . The path difference between waves reaching the observation point from two consecutive zones is: Δ = λ/2 Therefore, the waves from successive zones reach the observation point with a phase difference of...

  Diffraction of Light Diffraction is the phenomenon of the bending and spreading of light waves around the edges of an obstacle or through a narrow aperture when the size of the obstacle or aperture is comparable to the wavelength of light. It is a clear demonstration of the wave nature of light and occurs due to the superposition and interference of secondary wavelets . Types of Diffraction Diffraction is mainly of two types: 1. Fresnel Diffraction In Fresnel diffraction, the source and the screen are at finite distances from the diffracting obstacle or aperture. The incident wavefront is generally spherical or cylindrical . Examples: Diffraction by a straight edge Diffraction by a narrow slit Diffraction by a circular aperture 2. Fraunhofer Diffraction In Fraunhofer diffraction, the source and the screen are effectively at infinite distances from the diffracting aperture or obstacle. In practice, lenses are used to produce parallel incident and diffracted wavef...

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: \[ ...

Michelson's Interferometer 1. Construction Michelson's interferometer is an optical instrument used to produce interference fringes by dividing a single beam of light into two coherent beams and then recombining them.   Its main components are: Monochromatic Light Source (S): Provides light of a single wavelength. Beam Splitter (G₁): A semi-silvered glass plate placed at 45° to the incident beam. It divides the incident light into two parts. Compensating Plate (G₂): A plane-parallel glass plate of the same material and thickness as the beam splitter. It ensures that both beams pass through equal thicknesses of glass. Mirrors M₁ and M₂: Plane mirrors placed perpendicular to each other. One mir...

Interference by transmitted rays Consider a plane-parallel transparent thin film of refractive index μ and thickness t . A ray of monochromatic light is incident on the upper surface of the film at an angle of incidence i . A part of the incident ray is reflected from the upper surface, while the remaining part is refracted into the film. let draw two perpendicular $C_2B$ and $P_2M$. then effective pathdifference $$\Delta_1 = \mu(C_1P_2 - P_2C_2)- C_1B \qquad ...(1)$$ $\because \Delta C_1MP_2 \approx \Delta C_2MP_2$ $$\therefore C_1P_2 = P_2C_2 = \frac{t}{\cos r} \qquad ...(2)$$ $$ and \qquad C_1M = MC_2 = t\tan r \qquad ...(3)$$ In $\Delta C_1AC_2$ $$\sin i = \frac{C_1A}{C_1C_2}= \frac{C_1A}{C_1N + NC_2}$$ Hence, $$C_1A = 2t \sin i \tan r \qquad..(4) \qquad \because eq.(3)$$ Thus by...

Interference in Reflected Rays Interference in Reflected Rays Consider a plane-parallel transparent thin film of refractive index μ and thickness t . A ray of monochromatic light is incident on the upper surface of the film at an angle of incidence i . A part of the incident ray is reflected from the upper surface, while the remaining part is refracted into the film. The refracted ray travels inside the film and make refraction angle r . It is partially reflected from the lower surface. The two reflected rays subsequently emerge in the same direction and interfere with each other. The interference between these two reflected rays depends upon the optical path difference between them. let draw two perpendicular $P_2A$ and $C_1M$. then effective pathdifference $$\Delta_1 = \mu(P_1C_1 - C_1P_2)- P_1A \qquad ...(1)$$ $\bec...

  Plane-parallel thin film A monochromatic light ray S S  is incident obliquely on the upper surface of the film at P 1 P_1 ​ with an angle of incidence i i . At P 1 P_1 , the incident ray is divided into a reflected ray R 1 R_1  and a refracted ray that enters the film at an angle r r . The refracted ray travels through the film and reaches the lower surface at C 1 C_1 , where it is again divided into a transmitted ray T 1 T_1 ​ and a reflected ray. This reflected ray travels upward to the upper surface at P 2 P_2 ​ , where a part emerges as the second reflected ray R 2 R_2 ​ , while the remaining part is reflected again towards the lower surface. In the same manner, repeated reflections and transmissions take place between the two parallel surfaces of the film, producing successive reflected rays R 1 , R 2 , R 3 , … R_1, R_2, R_3,\ldots  and transmitted rays T 1 , T 2 , … T_1,T_2,\ldots . Since all these emergent rays are parallel and originate from the same inci...

  The complete experimental arrangement for determining the wavelength of monochromatic light with the help of Fresnel’s biprism is shown in Figure. Suppose the light rays are deviated through an angle δ by the refracting faces of Fresnel’s biprism. If the angle of each prism is α and its refractive index is μ, then the angle of deviation is: δ = α(μ − 1) If the distance between the slit S and the biprism is a, then, according to Fig., d ⁄ 2 = a tan δ Since δ is very small, tan δ = δ Therefore, d ⁄ 2 = aδ Substituting the value of δ, d ⁄ 2 = aα(μ − 1) or, d = 2aα(μ − 1) If the distance between the biprism and the eyepiece is b, then the distance between the slit S and the eyepiece is: D = (a + b) If the width of the interference fringes is β, then: β = λD ⁄ d Therefore, λ = βd ⁄ D Substituting the values of d and D, λ = β[2aα(μ − 1)] ⁄ (a + b) Thus, the wavelength of monochromatic light can be determined using Fresnel’s biprism.

In this experiment, the thickness of a thin film is determined with the help of Fresnel’s biprism. Suppose the central bright fringe, which is initially formed at P 0 , is observed on the screen. Let the distance between the virtual coherent sources S 1 and S 2 be d , and the distance between the virtual sources and the screen be D . A transparent thin film of thickness t and refractive index μ is introduced in the path of one of the interfering beams. Due to the introduction of the film, the central bright fringe is displaced from P 0 to P 1 through a distance x. The optical path difference produced due to the thin film is: (μt − t) = t(μ − 1) The path difference corresponding to the displacement x of the fringe is: xd ⁄ D Therefore, ...

  Fresnel’s biprism is an optical device used to produce two coherent virtual sources from a single monochromatic source. It is used to demonstrate the interference of light and to determine the wavelength of monochromatic light . A Fresnel biprism consists of two thin prisms joined at their bases. It is equivalent to two thin prisms placed base-to-base. Structurally, the biprism acts as a single piece of optical glass with one obtuse angle (approx. 179°) and two small base angles (approx. 0.5°). When light from a narrow slit S  falls on the biprism: The upper half deviates the light in one direction. The lower half deviates the light in the opposite direction. The rays appear to come from two virtual images, S 1 ​ and S 2 ​ , of the original source S . Thus, S 1 ​ and S 2 ​ act as two coherent sources and produce interference fringes on the screen.

  There are two different methods of producing coherent sources: By division of the wavefront  Young's Double-Slit Experiment : A single source of monochromatic light illuminates a barrier with two closely spaced slits. The light emerging from these two slits acts as coherent sources. Fresnel's Biprism Method : A single wavefront is refracted through two adjacent, acute-angled prisms to produce two coherent virtual sources By division of the amplitude Newton's Rings: Light reflects back and forth between a spherical lens and a flat glass plate, dividing the amplitude to produce a coherent interference pattern.  Michelson's Interferometer: A beam splitter divides the amplitude of a light beam into two paths, which are then reflected by mirrors and recombined to create interference.