/*! 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 82 A transparent rod \(30.0 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 the rod is viewed from its flat end, the apparent depth of the object is \(8.20 \mathrm{~cm}\) from the flat end. What is its apparent depth when the rod is viewed from its curved end?

Short Answer

Expert verified
The apparent depth when the rod is viewed from its curved end is \(10 cm + |d_{2}|\) (from Step 3).

Step by step solution

01

Apply Snell's Law

When viewing from the flat end, the apparent depth of 8.20 cm is given which is less than actual depth of 15 cm. Apply Snell's Law which states that \( n_{1}d_{1} = n_{2}d_{2} \). Here \(n_{1}\) is the refractive index of the air (1), \(d_{1}\) is the actual depth (15 cm), \(n_{2}\) is the refractive index of the rod (unknown) and \(d_{2}\) is the apparent depth (8.20 cm). From this, solve for \(n_{2}\) (refractive index of the rod).
02

Calculate Apparent Depth when Viewing from the Curved End

Now that we have found the refractive index of the rod, apply Snell's Law once again. This time, apply it on the curved end which has the radius of 10 cm. First, find the actual depth from the vertex to the embedded object (which is the radius minus half the rod's length, i.e., \(10 cm - 15 cm = -5 cm\)). The apparent depth from the curved end can be found by \( d_{2} = \frac{n_{1}d_{1}}{n_{2}} \).
03

Adjust Reading for Curved End

Since negative value for depth is not physically relevant, the negative sign indicates that the image is not within the curved end but is beyond this end on the other side. Therefore, the apparent depth when viewed from the curved end is the magnitude of the result obtained from Step 2, added to the radius of the curved end (10 cm), resulting in our final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Refractive Index
The refractive index, often denoted as 'n', is a measure of how much light bends, or refracts, when entering a material from another medium. Typically, the refractive index is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. The value indicates how much slower light travels in the material compared to a vacuum.

When we say that the refractive index of air is approximately 1, we imply that light travels in air at nearly the same speed as in a vacuum. If a material has a refractive index greater than 1, such as glass or water, it means that light slows down when it enters that material. This slowing down effects the bending of light rays at the interface, a fundamental aspect of optics covered by Snell's Law.

In the given exercise, the refractive index of the rod is not initially known, but it can be calculated using the provided data on apparent depth and actual depth. Knowing the refractive index is critical for understanding how light will behave as it passes through different parts of the rod, both flat and curved.
Apparent Depth
Apparent depth is an optical illusion that occurs when you observe an object submerged in a medium such as water or glass. It's the depth at which the object appears to be located due to the refraction of light as it exits the medium and enters your eyes. This effect is responsible for the 'bent stick' illusion seen when a stick is partially submerged in water—it looks bent or broken at the water surface, even though the stick is straight.

In our exercise context, it's mentioned that when observed from the flat end of the rod, the object has an apparent depth different from its actual position. This variation is due to the refraction of light within the rod, and it's paramount in calculating the refractive index of the rod. Understanding and calculating apparent depth, especially in materials of varying shapes, is essential to describe accurately how light interacts with objects embedded within or behind such materials.
Refraction of Light
Refraction is the change in direction of a wave, such as a light wave, when it passes from one medium to another at an angle. Snell's Law quantifies this behavior by relating the angle of incidence to the angle of refraction, affected by the refractive indices of the two media. The formula for Snell's Law is: \[ n_1 \cdot \sin(\theta_1) = n_2 \cdot \sin(\theta_2) \],where \(n_1\) and \(n_2\) are the refractive indices of the first and second media respectively, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction.

In our scenario, the light carrying information about the object's position within the rod changes direction as it exits the rod into the air, causing the object to appear at a different depth than where it is actually located. This phenomenon is not only fascinating but also has practical applications such as in lens design, correcting vision with glasses, and even in fabricating optical illusions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) You want to use a lens with a focal length of \(35.0 \mathrm{~cm}\) to produce a real image of an object, with the height of the image twice the height of the object. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

A thin lens with a focal length of \(6.00 \mathrm{~cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, \(25.0 \mathrm{~cm}\) from the eye, and that the lens is very close to the eye.

A convex spherical mirror with a focal length of magnitude \(24.0 \mathrm{~cm}\) is placed \(20.0 \mathrm{~cm}\) to the left of a plane mirror. An object \(0.250 \mathrm{~cm}\) tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

A pencil that is \(9.0 \mathrm{~cm}\) long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead \(12.0 \mathrm{~cm}\) from the mirror surface and the end of the eraser \(21.0 \mathrm{~cm}\) from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?

A concave mirror is to form an image of the filament of a headlight lamp on a screen \(8.00 \mathrm{~m}\) from the mirror. The filament is \(6.00 \mathrm{~mm}\) tall, and the image is to be \(24.0 \mathrm{~cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.