/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 A transparent rod 30.0 \(\mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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A transparent rod 30.0 \(\mathrm{cm}\) long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 \(\mathrm{cm}\) at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 \(\mathrm{cm}\) from the flat end and 15.0 \(\mathrm{cm}\) from the vertex of the curved end. When viewed from the flat end of the rod, the apparent depth of the object is 9.50 \(\mathrm{cm}\) from the flat end. What is its apparent depth when viewed from the curved end?

Short Answer

Expert verified
The apparent depth from the curved end is 8.02 cm.

Step by step solution

01

Understand the Apparent Depth Formula

The apparent depth for a flat surface can be calculated using the formula: \[D_{apparent} = \frac{D_{real}}{n}\]where \(D_{apparent}\) is the apparent depth, \(D_{real}\) is the real depth, and \(n\) is the refractive index of the rod material.
02

Calculate the Refractive Index

Given the apparent depth from the flat end is 9.50 cm, and the real depth from the flat end is 15.0 cm. Using the formula:\[9.50 = \frac{15.0}{n}\]Solving for \(n\), we multiply both sides by \(n\) and divide by 9.50 cm to get:\[n = \frac{15.0}{9.50} \approx 1.58\]
03

Use Lens Formula for the Curved Surface

For the curved end, we use the lens maker's formula for apparent depth:\[D_{apparent, curved} = \frac{R \cdot D_{real}}{(R - D_{real}) + n \cdot D_{real}}\]where \(R\) is the radius of curvature and \(D_{real}\) is the distance from the object to the vertex of the surface.
04

Substitute Known Values and Solve

Here, \(R = 10.0\) cm, \(D_{real} = 15.0\) cm, and \(n = 1.58\). Substitute these values into the equation:\[D_{apparent, curved} = \frac{10.0 \cdot 15.0}{(10.0 - 15.0) + 1.58 \cdot 15.0}\]Calculate the denominator:\[(10.0 - 15.0) + 1.58 \cdot 15.0 = -5.0 + 23.7 = 18.7\]So,\[D_{apparent, curved} = \frac{150.0}{18.7} \approx 8.02 \text{ cm}\]
05

Conclusion

The apparent depth of the object when viewed from the curved end of the rod is approximately 8.02 cm.

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Key Concepts

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

Refractive Index
The refractive index is a fundamental property of materials that describes how light propagates through them. Simply put, it tells us how much slower light travels in the material compared to a vacuum. This is important when dealing with optical devices like lenses or transparent rods, as it affects how we perceive the position of objects behind these materials. Specifically, for any material, the refractive index can be calculated using the formula:
  • Refractive Index (\(n\)):\(n = \frac{c}{v}\)
where\(c\)is the speed of light in a vacuum and\(v\)is the speed of light in the material. A higher refractive index indicates that light slows down more when entering the substance.

In our problem, we used the refractive index to find out how the depth of an object changes when viewed through a transparent rod. By comparing the real depth (how deep the object truly is) with the apparent depth (how deep it appears), we can use the formula:
  • \(D_{apparent} = \frac{D_{real}}{n}\)
This equation shows that the apparent depth is less than the real depth, due to the refractive property's nature of the rod.
Lens Maker's Formula
The lens maker's formula is a useful equation in optics, particularly when dealing with lenses that have curved surfaces. It allows us to calculate the apparent depth or position of an object seen through such lenses. This formula is slightly more complex than the one for a flat surface because it accounts for curvature. The lens maker’s formula for apparent depth through a curved surface is as follows:
  • \(D_{apparent, curved} = \frac{R \cdot D_{real}}{(R - D_{real}) + n \cdot D_{real}}\)
where,
  • \(R\)is the radius of curvature of the lens,
  • \(D_{real}\)is the real depth or distance from the object to the lens's vertex,
  • and\(n\)is the refractive index.
This formula helps determine how a curved lens affects the image location. In our problem, understanding this formula was key to determining the object’s apparent depth when viewed from the curved end of the rod.
Radius of Curvature
The radius of curvature is a critical concept in lens and optics. It describes the radius of the hypothetical sphere from which a curved surface or lens is a part. Understanding the radius of curvature helps us understand how strongly the lens will converge or diverge light.

In optics, the power of a lens and the way it manipulates light largely depend on this radius. A smaller radius of curvature implies a stronger bending of light, while a larger radius indicates a gentler curve.
  • When applied to the problem, a radius of 10.0 cm was used to represent the curvature of the rod's hemispherical surface.
In conjunction with the lens maker's formula, the radius of curvature helped calculate the object’s apparent depth as it appears through the curved end of the rod. Thus, knowing the radius is crucial for applying the formula accurately and predicting the optical behavior of lenses and curved surfaces.

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Most popular questions from this chapter

A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a \(4.00-\mathrm{mm}\)-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is \(6.00 \mathrm{cm},\) and the radius of curvature at the right end is 12.0 \(\mathrm{cm} .\) The length of the rod between vertices is 40.0 \(\mathrm{cm} .\) The object for the surface at the left end is an arrow that lies 23.0 \(\mathrm{cm}\) to the left of the vertex of this surface. The arrow is 1.50 \(\mathrm{mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 \(\mathrm{m}\) to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

A person swimming 0.80 \(\mathrm{m}\) below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 \(\mathrm{m}\) above the swimmer. What is the actual height of the diving board above the surface of the water?

Focus of the Eye. The cornea of the eye has a radius of curvature of approximately \(0.50 \mathrm{cm},\) and the aqueous humor behind it has an index of refraction of \(1.35 .\) The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 \(\mathrm{mm} .\) (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 \(\mathrm{cm}\) in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about \(5.0 \mathrm{mm},\) where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

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