Ch. 34 - The Wave Nature of Light: Interference and Polarization

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 34 - The Wave Nature of Light: Interference and Polarization

Problem 37

Chapter 33, Problem 37

Show that the radius r of the mᵗʰ dark Newton’s ring, as viewed from directly above (Fig. 34–18), is given by r = √mλR where R is the radius of curvature of the curved glass surface and λ is the wavelength of light used. Assume that the thickness of the air gap is much less than R at all points and that r ≪ R . [Hint: Use the binomial expansion.]

Verified step by step guidance

Start by understanding the setup: Newton's rings are formed due to the interference of light reflected from the top and bottom surfaces of the thin air film between a curved glass surface and a flat glass plate. The condition for destructive interference (dark rings) is that the optical path difference equals an odd multiple of half the wavelength.

The optical path difference for the mᵗʰ dark ring is given by: 2t = (m + 1/2)λ, where t is the thickness of the air film at the radius r, λ is the wavelength of light, and m is the ring number (m = 0, 1, 2, ...).

For small thickness t, the geometry of the curved surface gives the relationship: t ≈ r² / (2R), where r is the radius of the ring and R is the radius of curvature of the curved glass surface. Substitute this expression for t into the interference condition.

After substitution, the equation becomes: 2(r² / (2R)) = (m + 1/2)λ. Simplify this equation to isolate r², which gives: r² = mλR (neglecting the +1/2 term for large m, as it becomes negligible).

Finally, take the square root of both sides to solve for r: r = √(mλR). This is the desired expression for the radius of the mᵗʰ dark Newton's ring.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Rings

Newton's rings are a series of concentric circular interference patterns created by the reflection of light between a flat glass plate and a convex lens. The rings result from the varying thickness of the air gap between the two surfaces, leading to constructive and destructive interference of light waves. The radius of these rings can be mathematically related to the wavelength of light and the curvature of the lens.

Interference of Light

Interference occurs when two or more light waves overlap and combine, resulting in a new wave pattern. This phenomenon can be constructive, where waves reinforce each other, or destructive, where they cancel each other out. The conditions for interference depend on the path difference between the waves, which is crucial for understanding the formation of Newton's rings.

Binomial Expansion

The binomial expansion is a mathematical technique used to expand expressions that are raised to a power, particularly useful when dealing with approximations. In the context of Newton's rings, it allows for simplifying expressions involving the radius of curvature and the small thickness of the air gap. This simplification is essential for deriving the relationship between the radius of the mᵗʰ dark ring, the wavelength of light, and the radius of curvature.

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Related Practice

Textbook Question

A single optical coating reduces reflection to zero for λ = 550 nm. By what factor is the intensity reduced by the coating for λ = 430 nm and λ = 670 nm as compared to no coating? Assume normal incidence.

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Textbook Question

A uniform thin film of alcohol (n = 1.36) lies on a flat glass plate (n = 1.56). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for λ = 492 nm and a maximum for λ = 615 nm. What is the minimum thickness of the film?

Textbook Question

(I) If a soap bubble is 120 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated normally by white light? Assume that n = 1.35.

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Textbook Question

What is Brewster’s angle for an air-glass (n = 1.56) surface? Specify two answers.

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Textbook Question

(II) When a Newton’s ring apparatus (Fig. 34–18) is immersed in a liquid, the diameter of the tenth dark ring decreases from 2.92 cm to 2.54 cm. What is the refractive index of the liquid? [Hint: See Problem 37.]

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Textbook Question

Consider three equally spaced and equal-intensity coherent sources of light (such as adding a third slit to the two slits of Fig. 34–12). Determine the positions of minima.

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