/*! 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 15 A speck of dirt is embedded 3.50... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A speck of dirt is embedded 3.50 \(\mathrm{cm}\) below the surface of a sheet of ice \((n=1.309) .\) What is its apparent depth when viewed at normal incidence?

Short Answer

Expert verified
The apparent depth is approximately 2.67 cm.

Step by step solution

01

Understand the Problem Context

We are given a speck of dirt embedded 3.5 cm below the surface of ice. The refractive index of ice is given as 1.309. Our task is to find the apparent depth when viewed normally (perpendicularly) to the surface.
02

Recall the Apparent Depth Formula

The apparent depth can be calculated from real depth using the formula: \( \text{Apparent Depth} = \frac{\text{Real Depth}}{n} \), where \(n\) is the refractive index of the medium. This formula arises due to the bending of light as it passes from one medium to another.
03

Substitute Given Values into the Formula

Substitute the values into the formula. Real Depth is 3.5 cm and \(n = 1.309\). So, the equation becomes: \[ \text{Apparent Depth} = \frac{3.50}{1.309} \].
04

Calculate the Apparent Depth

Carry out the division to find the apparent depth: \[ \text{Apparent Depth} = \frac{3.50}{1.309} \approx 2.67 \text{ cm} \].
05

Verify and Conclude

Verify that the units are consistent and derived using the correct method. The result makes intuitive sense because the apparent depth should be less than the real depth when viewed from air into a denser medium like ice.

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 concept of refractive index is a fundamental characteristic of optical materials. It illustrates how much light bends, or refracts, as it enters or exits a material from another. The refractive index is represented by the symbol \( n \).

Mathematically, it is the ratio of the speed of light in a vacuum to its speed in the given medium. Therefore, the formula is given by:

\[ n = \frac{c}{v} \]

where:
  • \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).
  • \( v \) is the speed of light in the medium.

In practical terms, this means that a medium with a high refractive index slows down light significantly more than a medium with a low refractive index. In our example, ice has a refractive index of 1.309, indicating that light travels slower in ice than in air or vacuum. The refractive index plays a crucial role in calculating apparent depth and understanding light refraction.
Apparent Depth
When light moves from a denser medium to a less dense medium, like from water to air, the direction change makes objects appear closer than they actually are. This optical distortion is what we refer to as apparent depth.

Apparent depth determines how deep an object appears when viewed from above the surface of a transparent material. It is calculated using the formula:

\[ \text{Apparent Depth} = \frac{\text{Real Depth}}{n} \]

where:
  • \( \text{Real Depth} \) is the actual distance below the surface.
  • \( n \) is the refractive index of the material.

For instance, if you look at a stone embedded in ice, this formula indicates that the stone will seem shallower compared to its actual position beneath the ice. The refractive index causes the light from the stone to bend, creating the illusion of a less than real depth.
Light Bending
Light bending is a phenomenon extensively seen in everyday life and is scientifically known as refraction. This occurs when light travels through materials of different densities, changing speed and direction. It's what causes a straw to look bent in a glass of water.

The bending occurs because light speed varies in different media. In our exercise, light travels slower in ice compared to air, due to ice's higher refractive index. When light transitions from air to ice, it bends towards the normal. This bending can be described by Snell's Law, which states:

\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]

where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the two media.
  • \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.

This concept explains why objects under water often seem distorted or shifted from their real position. Understanding light bending is key to mastering optics and explains a variety of natural phenomena, from rainbows to the twinkling of stars.

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 person can see clearly up close but cannot focus on objects beyond 75.0 \(\mathrm{cm} .\) She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?

A converging lens with a focal length of 90.0 \(\mathrm{cm}\) forms an image of a 3.20 -cm-tall real object that is to the left of the lens. The image is 4.50 \(\mathrm{cm}\) tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

A lens forms an image of an object. The object is 16.0 \(\mathrm{cm}\) from the lens. The image is 12.0 \(\mathrm{cm}\) from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 \(\mathrm{mm}\) tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

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.

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.