Osm Physics by Ashish

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...