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A double-convex thin lens has surfaces with equal radii of curvature of magnitude \(2.50 \mathrm{cm} .\) Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 \(\mathrm{cm}\) from the lens. What is the index of refraction of the lens?

Short Answer

Expert verified
The index of refraction of the lens is approximately 1.67.

Step by step solution

01

Understand the Lens Formula

The lens formula for a thin lens is \( \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \), where \( n \) is the index of refraction, \( f \) is the focal length of the lens, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. For a double-convex lens with equal radii, \( R_1 = R_2 = R \).
02

Assign Known Values

Given that the lens forms an image of a very distant object at \( 1.87 \) cm, this distance is the focal length \( f = 1.87 \) cm. The radii of curvature are given as \( R_1 = R_2 = 2.50 \) cm.
03

Simplify the Lens Formula

In the lens formula \( \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \), substitute \( R_1 = R_2 = R \). The formula simplifies to \( \frac{1}{f} = (n - 1)\left(\frac{1}{R} - \frac{1}{R}\right) = \frac{2(n - 1)}{R} \).
04

Substitute Known Values

Substitute the known values \( f = 1.87 \) cm and \( R = 2.50 \) cm into the simplified formula: \( \frac{1}{1.87} = \frac{2(n - 1)}{2.50} \).
05

Solve for the Index of Refraction

Rearrange the equation to solve for \( n \):\[ \frac{2.50}{1.87} = 2(n - 1) \].Calculate the left side:\[ \frac{2.50}{1.87} = 1.337 \].Therefore, \[ 1.337 = 2(n - 1) \].Finally, solve for \( n \):\[ 1.337 = 2n - 2 \].\[ 2n = 1.337 + 2 \].\[ n = \frac{3.337}{2} \].\[ n = 1.6685 \].
06

Conclusion

The index of refraction of the lens is \( n \approx 1.67 \).

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Key Concepts

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

Index of Refraction
The index of refraction, often symbolized as \( n \), is a crucial concept in optical physics. It measures how much light bends or "refracts" when entering a material from a vacuum (or air in most practical scenarios). The index of refraction can be thought of as the ratio of the speed of light in a vacuum to its speed in the given material. A higher index means that light travels slower in that medium.

To find the index of refraction, you can use the lens-maker's formula, which for our specific problem simplifies to \( \frac{1}{f} = \frac{2(n-1)}{R} \) due to equal radii of curvature. By substituting the known values of focal length and radii, we calculate the index to discern its optical path characteristics.
  • Key Idea: Represents light speed difference relative to vacuum.
  • Application: Helps in calculating how much a lens will bend light.
Double-Convex Lens
A double-convex lens is a lens design where both surfaces are outwardly curved. This curvature allows it to converge light to a focal point, making it quite useful for focusing applications in optics, like magnifying glasses, cameras, and even your eyes.

The lens in our example is double-convex with equal radii. This means when parallel rays of light hit the lens, they are bent inward to meet at a point—a behavior dictated by the lens's material and shape. These lenses are key in converging light, thanks to their symmetric structure, which allows uniform focusing.
  • Shape: Outwardly curved on both sides.
  • Function: Focuses incoming parallel light rays to a point.
Focal Length
The focal length of a lens is the distance from the lens to the point where it brings incoming parallel light rays to a focus. It is a crucial parameter that determines the lens's ability to magnify or minimize images.

For a double-convex lens, shorter focal lengths imply stronger converging power. In our given exercise, the focal length is provided as 1.87 cm—the distance necessary to form a sharp image of a distant object. Understanding this helps in maneuvering applications where precision in image formation is required.
  • Definition: Distance from lens to focused image point.
  • Importance: Influences magnification and clarity of images.
Radii of Curvature
The radii of curvature relate to how deeply curved the lens surfaces are. In simpler terms, imagine the curvature as part of a sphere— the radius of this sphere is known as the radius of curvature.

In our scenario, the double-convex lens has equal and known radii of 2.50 cm for both surfaces. These equal curvatures simplify calculations, allowing the direct application of the lens formula to solve for unknowns like the index of refraction.

Changing the radii significantly alters the lens's ability to bend light, thus impacting its focal length. This is why precise knowledge of radii is essential in lens-making and optical design.
  • Understanding: Relates to the shape depth of the lens surfaces.
  • Affect: Critical in defining lens characteristics such as focal strength.

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Most popular questions from this chapter

A layer of benzene \((n=1.50) 2.60 \mathrm{cm}\) deep floats on water \((n-1.33)\) that is 6.50 \(\mathrm{cm}\) deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence?

A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

One end of a long glass rod is ground to a convex hemispherical shape. This glass an index of refraction of \(1.55 .\) When a small leaf is placed 20.0 \(\mathrm{cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 \(\mathrm{cm}\) from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33\()\) but nothing else were changed?

A symmetric, double-convex, thin lens made of glass with index of refraction 1.52 has a focal length in air of 40.0 \(\mathrm{cm}\) . The lens is sealed into an opening in the left hand end of a tank filled with water. At the right-hand end of the tank, opposite the lens, is a plane mirror 90.0 \(\mathrm{cm}\) from the lens. The index of refraction of the water is \(\frac{4}{3}\) . (a) Find the position of the image formed by the lens-water-mirror system of a small object outside the tank on the lens axis and 70.0 \(\mathrm{cm}\) to the left of the lens. (b) Is the image real or virtual? (c) Is it erect or inverted? (d) If the object has a height of \(4.00 \mathrm{mm},\) what is the height of the image?

Virtual Object. If the light incident from the left onto a convex mirror does not diverge from an object point but instead converges toward a point at a (negative) distance \(s\) to the right of the mirror, this point is called a virtual object. (a) For a convex mirror having a radius of curvature of \(24.0 \mathrm{cm},\) for what range of virtual-object positions is a real image formed? (b) What is the orientation of this real image? (c) Draw a principal- ray diagram showing the formation of such an image.

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