Interference in Thin Films

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Questions and Answers

In thin film interference, what path difference results in constructive interference (bright film) in reflected light, considering a phase change of $\lambda/2$?

  • $2 \mu t \cos r = n \lambda$
  • $2 \mu t \cos r = (2n - 1) \lambda / 2$
  • $2 \mu t \cos r = (2n + 1) \lambda / 2$ (correct)
  • $2 \mu t \cos r = (n + 1) \lambda$

If a thin film with refractive index $\mu$ and thickness $t$ is illuminated at an angle $r$, what is the optical path difference between two light rays reflected from the top and bottom surfaces, excluding phase change?

  • $2 \mu t \cos r$ (correct)
  • $2 \mu t$
  • $2t \cos r$
  • $ \mu t \cos r$

In Newton's rings experiment, why is the center of the ring pattern dark when viewed in reflected light?

  • Due to constructive interference at the center.
  • Because the thickness of the air film is equal to $\lambda$ at the center.
  • Due to destructive interference and a path difference of $\lambda/2$. (correct)
  • Due to destructive interference and a path difference of $\lambda/4$.

In the context of Newton's rings, what does the radius of the $n^{th}$ dark ring depend on?

<p>Both the wavelength of light and the radius of curvature of the lens. (B)</p> Signup and view all the answers

For Newton's rings formed by reflected light, how are the diameters of the dark rings related to natural numbers?

<p>Proportional to the square roots of natural numbers. (A)</p> Signup and view all the answers

What happens to the fringe width in Newton's rings as the order of the fringe increases?

<p>The fringe width decreases. (C)</p> Signup and view all the answers

In the experiment to determine the wavelength of sodium light using Newton's rings, what parameters are directly measured to calculate the wavelength?

<p>Diameters of dark rings only. (A)</p> Signup and view all the answers

During the determination of the refractive index of a liquid using Newton's rings, how does introducing the liquid affect the ring diameters?

<p>Decreases the diameters. (D)</p> Signup and view all the answers

What is the primary difference between interference due to division of wavefront and division of amplitude?

<p>Division of wavefront involves dividing the incident wavefront into two parts, while division of amplitude divides the amplitude of the incoming beam. (A)</p> Signup and view all the answers

In Fresnel's biprism experiment, what is the function of the biprism?

<p>To create two virtual coherent sources. (C)</p> Signup and view all the answers

In the experimental setup for Fresnel's biprism, what adjustments are crucial for observing clear interference fringes?

<p>Correctly leveling the optical bench and aligning the components. (C)</p> Signup and view all the answers

What is the effect of lateral shift in Fresnel's biprism experiment, and how is it corrected?

<p>It shifts the fringes sideways and is corrected by ensuring the slit and biprism are parallel to the optical bench. (B)</p> Signup and view all the answers

In Fresnel's biprism experiment, what does the fringe width, $\beta$, represent?

<p>The spacing between successive maxima or minima. (B)</p> Signup and view all the answers

Which of the following is NOT a required measurement when determining the wavelength of light using Fresnel's biprism?

<p>The refractive index of the biprism. (B)</p> Signup and view all the answers

In thin-film interference, what condition changes when the film is observed at an angle instead of head-on?

<p>The angle of refraction $r$. (C)</p> Signup and view all the answers

How does coating lenses with thin films reduce reflections?

<p>By causing destructive interference of reflected light waves. (A)</p> Signup and view all the answers

What is the effect of using white light in Newton's rings experiment?

<p>Colored rings are observed. (D)</p> Signup and view all the answers

When determining the radius of curvature of a plano-convex lens using a spherometer, what measurements are required?

<p>Both the distance between the legs and the difference in spherometer readings. (C)</p> Signup and view all the answers

In transmitted light Newton's rings, what condition applies for bright rings?

<p>The path difference is an integer multiple of the wavelength. (D)</p> Signup and view all the answers

How do the diameters of the bright rings change when Newton's Rings apparatus is immersed in a liquid of refractive index $\mu$?

<p>Decrease by a factor of $\sqrt{\mu}$. (B)</p> Signup and view all the answers

Flashcards

Path difference in thin films

The path difference between two reflected light rays in a thin film is given by (Δ = μ(BC + CD) - BE), where (μ) is the refractive index of the film, BC and CD are the paths in the film, and BE is the path in air.

Condition for Brightness

Maxima occur in reflected light when (2μt cos r = (2n + 1)λ/2), where (μ) is the refractive index, (t) is the thickness of the film, (r) is the angle of refraction, (n) is an integer, and (λ) is the wavelength of light.

Condition for Darkness

Minima occur in reflected light when (2μt cos r = nλ), where (μ) is the refractive index, (t) is the thickness, (r) is the angle of refraction, (n) is an integer, and (λ) is the wavelength of light.

Newton's Rings

Alternate dark and bright rings concentric around the point of contact between the lens and glass plate.

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Central Spot

At the point of contact in Newton's rings, the path difference is (λ/2), leading to destructive interference and a dark central spot.

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Division of Wavefront

Interference due to the division of the incident wavefront into two parts that travel unequal distances and reunite at an angle.

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Division of Amplitude

Interference where the incoming beam's amplitude is divided into two parts by reflection or refraction, then reunited after traversing different paths.

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Fringe Width

The fringe width is the spacing between successive interference maxima or minima. Given by (β = λD / 2d).

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Biprism Adjustments

Adjustments include leveling the optical bench, aligning the slit, biprism, and eyepiece, and ensuring proper illumination.

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Wavelength Determination

The wavelength of light in a biprism experiment is determined by (λ = β(2d) / D), where (β) is the fringe width, (2d)is the distance between virtual sources and (D) is the distance between the slit and eyepiece.

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Study Notes

  • Interference occurs when light waves combine, leading to phenomena like bright and dark fringes

Interference in Thin Films

  • Interference in thin films arises from light reflecting off the top and bottom surfaces of the film
  • The effective path difference decides the type of interference
  • Phase change of Ï€ (equivalent to λ/2 path difference) happens when light reflects from a denser medium's surface
  • Effective path difference between reflected rays is (2µt cos r±λ/2), where:
    • µ is refractive index
    • t is film thickness
    • r is angle of refraction

Constructive Interference

  • For maxima (bright reflection), the condition is 2µt cos r = (2n + 1) λ/2, where n is an integer

Destructive Interference

  • For minima (dark reflection), the condition is 2µt cos r = nλ

Newton's Rings Setup

  • Newton's rings are formed by placing a plano-convex lens on a glass plate, creating an air film of increasing thickness
  • When monochromatic light falls on the lens, a pattern of concentric dark and bright rings appears in reflected light
  • The rings are due to the interference between light waves reflected from the top and bottom surfaces of the air film

Theory of Newton's Rings (Reflected Light)

  • For bright rings, 2t = (2n - 1) λ/2
  • For dark rings, 2t = nλ, where n is an integer
  • Radius r of the Newton's ring is related to the film thickness via r² = 2Rt, where R is the radius of curvature
  • Diameter D of the nth bright ring is D = √(2λR(2n - 1))
  • Diameters of bright rings are proportional to the square roots of odd natural numbers
  • Diameter D of the nth dark ring is D = 2√(nλR)
  • Diameters of dark rings are proportional to the square roots of natural numbers

Fringe Width

  • Fringe width decreases as the order of the fringe increases

Transmitted Light

  • With transmitted light, the central ring is bright, which is the opposite of the rings in reflected light

Determining Wavelength

  • λ can be destermined as λ = (Dn+p² - Dn²)/4pR;
    • R is curvature radius
    • Dn and Dn+p are nth and (n+p)th ring diameters

Determining Refractive Index of Liquid

  • The experiment is repeated with liquid between lens & plate. You simply solve for µ
  • µ = (Dn+p² - Dn²)/(D'n+p² - D'n²), where D are radius without liquid and D' with liquid

Types of Interference:

  • Division of wavefront: Wavefront splits via reflection, refraction, or diffraction (ex: Fresnel Biprism)
  • Division of amplitude: Amplitude splits via reflection or refraction (ex: Newton's Rings)

Fresnel's Biprism:

  • Biprism creates two coherent images from a single slit. The images act as coherent sources

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