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A Newton's ring apparatus consists of a convex lens with a large radius of curvature \(R\) placed on a flat glass disc. (a) Show that the distance \(x\) from the center to the air, thickness \(d,\) and the radius of curvature \(R\) are given by \(x^{2}=2 R d\) (b) Show that the radius of nth constructive interference is given by \(x_{\mathrm{n}}=\left[\left(n+\frac{1}{2}\right) \lambda R\right]^{1 / 2} .\) (c) How many bright fringes may be seen if it is viewed by red light of wavelength 700\. \(\mathrm{nm}\) for \(R=10.0 \mathrm{~m},\) and the plane glass disc diameter is \(5.00 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: 24 bright fringes will be visible under the given conditions.

Step by step solution

01

Part (a): Deriving the relationship between x, R, and d

Start by drawing a straight line from the center of the circle of curvature to the point where the air thickness changes from d to 0. It meets the air film at point A and the surface AB at point B. Since both points A and B are on the circle with radius R, triangles AOB and ADB are both right-angled triangles. Using the Pythagorean theorem: \(R^2 = (R - d)^2 + x^2\) Expanding and simplifying, we get: \(x^2 = 2Rd\)
02

Part (b): Determining the relationship between the radius of the nth constructive interference and x

To find the expression for the radius \(x_n\) for the nth constructive interference, first, write down the condition for constructive interference: \(m\lambda = 2d\) Now, we need to substitute the expression for d that we derived in part (a). Rearrange the equation for d: \(d = \frac{x^2}{2R}\) Substitute this back into the constructive interference condition: \(m\lambda = 2\left(\frac{x^2}{2R}\right)\) Solve for x: \(x = \sqrt{(m \lambda R)}\) Since m = n + 1/2 for bright fringes, we can write the expression for the radius of the nth constructive interference as: \(x_n = \sqrt{\left(n+\frac{1}{2}\right) \lambda R}\)
03

Part (c): Calculating the number of bright fringes

Now that we have an expression for \(x_n\), we can calculate the number of visible bright fringes for the given values. The diameter of the glass disc is given as \(5.00 \: cm\), so the maximum radius is \(2.50 \: cm\). For R = 10.0 m, and the wavelength of red light as 700 nm, we have: \(x_n = \sqrt{\left(n+\frac{1}{2}\right) \lambda R} \leq 2.50 \: cm\) We need to find the largest integer value of n such that the above inequality holds. Rearrange the inequality for n: \(n \leq \frac{(2.50 \times 10^{-2}m)^2}{700 \times 10^{-9}m \times 10.0m} - \frac{1}{2}\) Calculate the maximum n value: \(n \leq \approx 24.91\) Because n must be an integer, we will choose 24 as the highest possible value for n. Therefore, 24 bright fringes can be seen for the given conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constructive Interference
Constructive interference is a phenomenon that occurs when two or more waves coincide with each other, resulting in a wave of greater amplitude. Imagine tossing two pebbles into a still pond close to each other. The ripples created by the pebbles would spread out and eventually overlap. Where the crests of the waves meet, they combine to form a wave with a higher peak; this is constructive interference, a principle that's fundamental in analyzing the pattern produced by Newton's rings.

Newton's rings themselves arise from the constructive interference of light waves reflecting off the top and bottom surfaces of a very thin air film created between a flat glass plate and a lens. The light waves that reflect from the lower surface travel a slightly longer path than those from the upper surface. When they reunite, their crests and troughs can line up to constructively interfere under the right conditions—those being a path difference between the waves that corresponds to an integer multiple of the wavelength of light (ewline).

In Newton's rings experiments, this alignment leads to concentric bright and dark fringes. The bright fringes correspond to the points of constructive interference where the path difference is a half-wavelength out of phase, allowing for the waves to sync up and amplify each other.
Radius of Curvature
The radius of curvature (ewline) is a measure of the degree of curvature of an object like a lens or mirror—in this case, specifically the convex lens used in the Newton's rings setup. Imagine a circle that perfectly aligns with the curved surface of the lens, the radius of that circle would be the lens's radius of curvature.

The relationship between the radius of curvature, the distance from the center (ewline), and the air film thickness (ewline) is pivotal in predicting the position of the bright and dark fringes seen in Newton's rings. It is established that ewline, highlighting the direct correlation between these elements. As you move further from the center of the pattern, the air gap increases and so does the order of the fringe.

The radius of curvature also significantly influences the size of the Newton's rings. A lens with a larger radius of curvature will produce rings with a larger radius for each successive fringe. This is essential to understanding the pattern's geometry and calculating the number of bright fringes visible, as seen in the exercise.
Wavelength of Light
The wavelength of light (ewline) is the distance between successive crests of a wave. It determines the color of the light and is measured in nanometers (nm). The visible light spectrum ranges from about 400 nm (violet) to 700 nm (red), with the given exercise dealing with red light at 700 nm.

The wavelength is a critical factor in the interference pattern observed in Newton's rings. It directly influences the spacing between the fringes—the constructive interference patterns appearing as bright rings. As the wavelength increases, the distance between successive bright fringes increases as well, which can be visualized by the relationship ewline for the radius of the nth bright fringe.

In practical terms, using light with a known wavelength allows one to calculate other properties, such as the radius of curvature of the lens or the index of refraction of the film, by analyzing the interference pattern. In the provided exercise, the wavelength of red light is used to determine the maximum number of observable bright fringes, giving us insight into how the physical properties of light interact with materials to create observable phenomena.

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Most popular questions from this chapter

The thermal stability of a Michelson interferometer can be improved by submerging it in water. Consider an interferometer that is submerged in water, measuring light from a monochromatic source that is in air. If the movable mirror moves a distance of \(d=0.200 \mathrm{~mm},\) exactly \(N=800\) fringes move by the screen. What is the original wavelength (in air) of the monochromatic light?

In a double-slit arrangement the slits are \(1.00 \cdot 10^{-5} \mathrm{~m}\) apart. If light with wavelength \(500 .\) nm passes through the slits, what will be the distance between the \(m=1\) and \(m=3\) maxima on a screen \(1.00 \mathrm{~m}\) away?

White light is shone on a very thin layer of mica \((n=1.57),\) and above the mica layer, interference maxima for two wavelengths (and no other in between) are seen: one blue wavelength of \(480 \mathrm{nm},\) and one yellow wavelength of \(560 \mathrm{nm} .\) What is the thickness of the mica layer?

A two-slit apparatus is covered with a red \((670 \mathrm{nm})\) filter. When white light is shone on the filter, on the screen beyond the two-slit apparatus, there are nine interference maxima within the 4.50 -cm-wide central diffraction maximum. When a blue \((450 \mathrm{nm})\) filter replaces the red, how many interference maxima will there be in the central diffraction maximum, and how wide will that diffraction maximum be?

A red laser pointer with a wavelength of \(635 \mathrm{nm}\) shines on a diffraction grating with 300 lines \(/ \mathrm{mm}\). A screen is then placed a distance of \(2.0 \mathrm{~m}\) behind the diffraction grating to observe the diffraction pattern. How far away from the central maximum will the next bright spot be on the screen? a) \(39 \mathrm{~cm}\) c) \(94 \mathrm{~cm}\) e) \(9.5 \mathrm{~m}\) b) \(76 \mathrm{~cm}\) d) \(4.2 \mathrm{~m}\)

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