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(II) A planoconvex lucite lens 3.4 \(\mathrm{cm}\) in diameter is placed on a flat piece of glass as in Fig. 18. When 580 -nm light is incident normally, 44 bright rings are observed, the last one right at the edge. What is the radius of curvature of the lens surface, and the focal length of the lens? [Hint: see Problem \(33 .\)]

Short Answer

Expert verified
Radius of curvature \( R \) is approximately 11.32 m, and focal length \( f \) is approximately 11.55 m.

Step by step solution

01

Understand the Equation for Bright Rings

For interference in thin films, specifically for Newton's rings formed by a planoconvex lens on a flat surface, the condition for constructive interference (bright rings) is given by the equation: \( m \cdot \lambda = 2 \cdot t_m \cdot n \), where \( m \) is the ring number, \( \lambda \) is the wavelength of light, \( t_m \) is the thickness of the air film at the m-th bright ring, and \( n \) is the refractive index of air (approximately 1).
02

Relate Thickness to Ring Radius

The thickness \( t_m \) for a planoconvex lens can be related to the radius \( r \) of the m-th bright ring by the geometry of the lens: \( t = \frac{r^2}{2R} \), where \( R \) is the radius of curvature of the lens, and \( r \) is the radius of the ring.
03

Substitute and Simplify the Equation

Combine the two equations, \( m \lambda = 2 \cdot \frac{r^2}{2R} \), to get \( m \cdot \lambda = \frac{r^2}{R} \), which simplifies to \( R = \frac{r^2}{m \cdot \lambda} \).
04

Calculate Radius of Curvature

Given that the diameter is 3.4 cm, the edge radius \( r = \frac{3.4}{2} \) cm = 1.7 cm = 0.017 m. The wavelength \( \lambda \) is 580 nm = 580 \( \times 10^{-9} \) m. With \( m = 44 \), substitute these values into \( R = \frac{r^2}{m \cdot \lambda} \):\[R = \frac{(0.017)^2}{44 \times 580 \times 10^{-9}} = \frac{0.000289}{0.02552} \approx 11.32 \text{ m}\]
05

Calculate the Focal Length

The focal length \( f \) of a lens is calculated using the lensmaker's equation for a lens in air: \( f = \frac{R}{2(n-1)} \), where \( n \) is the refractive index of lucite, typically around 1.49. Thus, \( f = \frac{11.32}{2 \times (1.49-1)} \approx \frac{11.32}{0.98} \approx 11.55 \text{ m}.\)

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Key Concepts

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

Newton's Rings
Newton's Rings are a fascinating phenomenon of light interference. They occur when a convex lens is placed on a flat glass surface. At the point where the two glass surfaces nearly touch, thin air layers are trapped. This subtle air gap varies in thickness, causing light waves to interfere—either constructively or destructively. As a result, you see a series of alternating bright and dark concentric rings when viewed under monochromatic light, like 580 nm light in this exercise.

Each bright ring represents a point of constructive interference. The order of the bright ring is important; for example, in the given exercise, 44 bright rings mean constructive interference at that particular level. This is crucial in calculating the Radius of Curvature (R) of the lens, using equations that relate the physical parameters to the observed interference patterns.
  • The first ring is called the zero-order ring.
  • Rings become fainter as you go outward.
Radius of Curvature
The Radius of Curvature (R) is a measure of how curved a surface is. It appears in optics, where you see it in concepts like Newton's Rings. This radius relates directly to the geometry of the lens and the light patterns you observe. In context, when light reflects between a lens and a flat surface, the curvature directly affects the thickness of the air film across which interference occurs.

Using the relationship \[ R = \frac{r^2}{m \cdot \lambda} \]where:
  • \( r \) is the radius of the m-th bright ring,
  • \( m \) is the number of the bright ring, and
  • \( \lambda \) is the wavelength of light,
you can calculate \( R \) from observed values. In our example, the radius of curvature was computed to be approximately 11.32 meters. This results from substituting in 44 bright rings and the known distance from lens center to edge, and wavelength.
Lensmaker's Equation
The Lensmaker's Equation provides a way to determine the focal length \( f \) of a lens, such as our lucite lens, using the Radius of Curvature and the refractive index of the lens material. Written as: \[ f = \frac{R}{2(n-1)} \]where:
  • \( R \) is the Radius of Curvature,
  • \( n \) is the refractive index of the material.
This equation assumes the lens is in air, which simplifies some calculations. For a lucite lens, you can get the focal length knowing the Radius of Curvature (11.32 meters here) and that lucite's refractive index is about 1.49.

The result gives us a focal length approximately equal to 11.55 meters, helpful in determining properties of lenses in optical devices, from glasses to microscopes.
Refractive Index
The Refractive Index (or index of refraction) is a dimensionless number that describes how light propagates through a medium. For any given material, the refractive index is crucial to understanding its optical properties. It determines how much the speed of light is reduced and how much light bends, or refracts, when entering that material.

For example, in lucite, the refractive index is around 1.49. This means that light travels slower in lucite compared to air, which has a refractive index close to 1. The refractive index helps calculate many properties, including the focal length of a lens, using the Lensmaker's Equation. It's a key parameter in designing lenses because it affects lens curvature and thickness.
  • Refractive index varies with wavelength (known as dispersion).
  • Higher refractive index means greater bending of light.
Understanding this concept is essential for manipulating light in various optical applications such as lenses and fiber optics.

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Most popular questions from this chapter

(II) In a double-slit experiment, the third-order maximum for light of wavelength \(500 \mathrm{nm}\) is located \(12 \mathrm{~mm}\) from the central bright spot on a screen \(1.6 \mathrm{~m}\) from the slits. Light of wavelength \(650 \mathrm{nm}\) is then projected through the same slits. How far from the central bright spot will the second-order maximum of this light be located?

Light of wavelength \(690 \mathrm{nm}\) passes through two narrow slits \(0.66 \mathrm{~mm}\) apart. The screen is \(1.60 \mathrm{~m}\) away. A second source of unknown wavelength produces its second-order fringe \(1.23 \mathrm{~mm}\) closer to the central maximum than the 690 -nm light. What is the wavelength of the unknown light?

Television and radio waves reflecting from mountains or airplanes can interfere with the direct signal from the station. (a) What kind of interference will occur when 75-MHz television signals arrive at a receiver directly from a distant station, and are reflected from a nearby airplane \(122 \mathrm{~m}\) directly above the receiver? Assume \(\frac{1}{2} \lambda\) change in phase of the signal upon reflection. (b) What kind of interference will occur if the plane is \(22 \mathrm{~m}\) closer to the receiver?

(II) A very thin sheet of plastic \((n=1.60)\) covers one slit of a double-slit apparatus illuminated by 680 -nm light. The center point on the screen, instead of being a maximum, is dark. What is the (minimum) thickness of the plastic?

(II) What is the wavelength of the light entering an interferometer if 384 bright fringes are counted when the movable mirror moves 0.125 \(\mathrm{mm}\) ?

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