/*! 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 1 A candle 4.85 \(\mathrm{cm}\) ta... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A candle 4.85 \(\mathrm{cm}\) tall is 39.2 \(\mathrm{cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

Short Answer

Expert verified
The image is formed 39.2 cm behind the mirror, and its height is 4.85 cm.

Step by step solution

01

Understanding Image Formation in Plane Mirrors

In plane mirrors, the image formed is a virtual image, meaning it appears to exist behind the mirror. The image is located at the same distance from the mirror as the object but on the opposite side.
02

Determine the Image Distance

Given that the object (candle) is 39.2 cm to the left of the mirror, the image will also be 39.2 cm from the mirror, but on the right side. This is because plane mirrors produce images that are equidistant from the object.
03

Determine the Height of the Image

In plane mirrors, the size of the image is the same as the size of the object. Since the candle is 4.85 cm tall, the height of the image will also be 4.85 cm.

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.

Understanding Virtual Images
In the world of optics, particularly when dealing with mirrors, we often hear about virtual images. A virtual image is an image that appears to be located at a position where light does not actually come from. When you look into a plane mirror, the reflection you see is a virtual image. This occurs because the light rays from an object do not actually meet behind the mirror to form the image. Instead, they only seem to diverge from a point behind the mirror.
- Virtual images are non-tangible because the light does not converge at their location. - They cannot be projected onto a screen like real images can. - Virtual images appear upright and are always the same size as the object. By understanding virtual images, you gain a better grasp of how reflections work in plane mirrors and why these images are significant in everyday observations.
Image Distance in Plane Mirrors
When discussing images in plane mirrors, an important concept is the image distance. Image distance refers to how far the image appears behind the mirror compared to the actual object in front of it. In a plane mirror, the image distance is always equal to the object distance. This means: - If an object is 39.2 cm in front of a mirror, the image will be 39.2 cm behind the mirror. - The image and object are equidistant from the mirror surface. Here’s why: plane mirrors create an illusion by reversing the direction of light. As the rays hit the mirror, they get reflected back at equal angles, forming a virtual image that appears the same distance behind the mirror as the object is in front. So, when you step 2 meters away, your image also appears 2 meters behind the mirror.
Image Height Calculation in Plane Mirrors
Calculating the height of an image formed by a plane mirror is straightforward and not intimidating at all. This is because plane mirrors maintain the size and proportions of the original object. In any plane mirror setup: - The image height will equal the object height. - If a candle is 4.85 cm tall, its image in the mirror will also be 4.85 cm tall. This rule is consistent because plane mirrors do not alter the scale of an image. All dimensions are preserved perfectly except for their virtual nature and lateral inversion. So, whenever you’re dealing with a plane mirror, just remember: the height of the image is a direct reflection of the height of the object!

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 camera lens has a focal length of 180.0 \(\mathrm{mm}\) and an aperture diameter of 16.36 \(\mathrm{mm}\) (a) What is the \(f\) -number of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{30} \mathrm{s}\) at \(f / 11\) , what is the correct exposure at \(f / 2.8 ?\)

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?

The focal length of the eyepiece of a certain microscope is 18.0 \(\mathrm{mm}\) . The focal length of the objective is 8.00 \(\mathrm{mm}\) . The distance between objective and eyepiece is \(19.7 \mathrm{cm} .\) The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

A camera lens has a focal length of 200 \(\mathrm{mm}\) . How far from the lens should the subject for the photo be if the lens is 20.4 \(\mathrm{cm}\) from the film?

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.