/*! 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 40 A drop of water on a countertop ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A drop of water on a countertop reflects light from a flower held \(3.0 \mathrm{cm}\) directly above it. The flower's diameter is \(2.0 \mathrm{cm},\) and the diameter of the flower's image is \(0.10 \mathrm{cm} .\) What is the focal length of the water drop, assuming that it may be treated as a convex spherical mirror?

Short Answer

Expert verified
The focal length of the water drop is approximately \(-0.158\, \text{cm}\).

Step by step solution

01

Identify Given Values

We are given that the object distance \(o\) is \(3.0\, \text{cm}\), the object diameter is \(2.0\, \text{cm}\), and the image diameter is \(0.10\, \text{cm}\).
02

Calculate the Magnification

The magnification \(m\) of a mirror is given by the ratio of the image diameter to the object diameter. Hence, \(m = \frac{0.10\, \text{cm}}{2.0\, \text{cm}} = 0.05\).
03

Relation Between Magnification and Image Distance

The magnification \(m\) is also equal to the negative ratio of the image distance \(i\) to the object distance \(o\), so \(m = -\frac{i}{o}\). Since \(m = 0.05\) and \(o = 3.0\, \text{cm}\), we have \(0.05 = -\frac{i}{3.0}\).
04

Solve for Image Distance

Rearrange the equation to solve for \(i\): \[i = -0.05 \times 3.0 = -0.15\, \text{cm}\].The negative sign indicates that the image is virtual, which is consistent with a convex mirror.
05

Use Mirror Equation to Find Focal Length

For a spherical mirror, the mirror equation is given by\[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]Substitute \(o = 3.0\, \text{cm}\) and \(i = -0.15\, \text{cm}\) into the equation. Hence:\[\frac{1}{f} = \frac{1}{3.0} + \frac{-1}{0.15}\]Calculate the values to find \(f\).
06

Calculate Focal Length

First, calculate the fractions: \[\frac{1}{3.0} = 0.3333\] \[\frac{1}{-0.15} = -6.6667\]Combine the fractions:\[\frac{1}{f} = 0.3333 - 6.6667 = -6.3334\]Therefore, the focal length is:\[f = \frac{1}{-6.3334} \approx -0.158\, \text{cm}\].The negative sign indicates a focal length consistent with a convex mirror.

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.

Focal Length
The focal length of a mirror is the distance from the mirror to its focal point. In the context of spherical mirrors, this focal length determines how light rays converge or appear to diverge from a point. In this exercise, we dealt with a convex spherical mirror formed by a water droplet. When light reflects off this mirror, the focal length tells us how the reflected light behaves relative to its surface. It is essential because it affects the size and orientation of the image formed by the mirror.
For convex mirrors, the focal length is generally considered negative. This is because convex mirrors cause light rays to diverge, and the virtual focus is located behind the mirror.
In the calculation provided, the determined focal length of approximately -0.158 cm is indicative of a virtual, diminished image typical of convex mirrors.
Convex Mirror
A convex mirror, or a diverging mirror, is characterized by its outwards curve, resembling the exterior of a sphere. Convex mirrors have the unique property of spreading out light rays that hit them, causing reflected light to diverge. This outward curvature results in certain distinctive features:
  • They form virtual images. A virtual image appears behind the mirror and cannot be projected onto a screen, as it is formed by the extensions of diverging rays.
  • These mirrors provide a wider field of view, making them popular for use in places like vehicle side mirrors and security applications.
  • They always produce images that are smaller than the actual object, which can be calculated using magnification and the mirror equation.
The water droplet in our exercise served as a convex mirror, illustrating how such surfaces manipulate light and affect both the image location and size.
Magnification
Magnification describes how much larger or smaller an image is compared to the actual object size. It is crucial for understanding how mirrors like the convex mirror in our exercise reshape the appearance of objects.In mathematical terms, magnification ( m ) is the ratio of the image size to the object size:\[m = \frac{\text{Image size}}{\text{Object size}}\]This parameter also links closely to the mirror equation, as it relates to the image ( i ) and object ( o ) distances:\[m = -\frac{i}{o}\]In our exercise, the water droplet as a convex mirror resulted in a magnification of 0.05, indicating a much smaller image compared to its actual size. A positive magnification would mean an upright image, while a negative value, such as what we computed, indicates an inverted image.
Mirror Equation
The mirror equation is a fundamental relation in optics, allowing us to connect the focal length, object distance, and image distance of mirrors, whether concave or convex.The equation is given by:\[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]where f is the focal length, o the object distance, and i the image distance.For our case involving a convex mirror, when substituting the known values into the mirror equation, we used this relationship to solve for the focal length of the water droplet. Because the resulting focal length was negative, it is consistent with the expectations for a convex mirror, emphasizing its virtual image formation capabilities.Utilizing the mirror equation helps in predicting and understanding how different mirrors affect light and image representation, as seen in the exercise.

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 concave mirror has a focal length of \(30.0 \mathrm{cm} .\) The distance between an object and its image is \(45.0 \mathrm{cm} .\) Find the object and image distances, assuming that (a) the object lies beyond the center of curvature and (b) the object lies between the focal point and the mirror.

A plane mirror and a concave mirror \((f=8.0 \mathrm{cm})\) are facing each other and are separated by a distance of \(20.0 \mathrm{cm} .\) An object is placed between the mirrors and is \(10.0 \mathrm{cm}\) from each mirror. Consider the light from the object that reflects first from the plane mirror and then from the concave mirror. Using a ray diagram drawn to scale, find the location of the image that this light produces in the concave mirror. Specify this distance relative to the concave mirror.

A dentist's mirror is placed \(2.0 \mathrm{cm}\) from a tooth. The enlarged image is located \(5.6 \mathrm{cm}\) behind the mirror. (a) What kind of mirror (plane, concave, or convex) is being used? (b) Determine the focal length of the mirror. (c) What is the magnification? (d) How is the image oriented relative to the object?

A small mirror is attached to a vertical wall, and it hangs a distance of \(1.80 \mathrm{m}\) above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the reflected ray strikes the floor at a distance of \(3.86 \mathrm{m}\) from the base of the wall. Later on in the morning, the ray is observed to strike the floor at a distance of \(1.26 \mathrm{m}\) from the wall. The earth rotates at a rate of \(15.0^{\circ}\) per hour. How much time (in hours) has elapsed between the two observations?

An object is placed in front of a convex mirror. Draw the convex mirror (radius of curvature \(=15 \mathrm{cm}\) ) to scale, and place the object \(25 \mathrm{cm}\) in front of it. Make the object height \(4 \mathrm{cm} .\) Using a ray diagram, locate the image and measure its height. Now move the object closer to the mirror, so the object distance is \(5 \mathrm{cm} .\) Again, locate its image using a ray diagram. As the object moves closer to the mirror, (a) does the magnitude of the image distance become larger or smaller, and (b) does the magnitude of the image height become larger or smaller? (c) What is the ratio of the image height when the object distance is \(5 \mathrm{cm}\) to its height when the object distance is \(25 \mathrm{cm} ?\) Give your answer to one significant figure.

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.