/*! 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 21 A moth at about eye level is \(1... [FREE SOLUTION] | 91Ó°ÊÓ

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A moth at about eye level is \(10 \mathrm{~cm}\) in front of a plane mirror; you are behind the moth, \(30 \mathrm{~cm}\) from the mirror. What is the distance between your eyes and the apparent position of the moth's image in the mirror?

Short Answer

Expert verified
40 cm

Step by step solution

01

Understand the Problem

The problem involves a mirror and reflections. A moth is in front of a plane mirror, and you are behind the moth. We need to determine the distance between your eyes and the apparent position of the moth’s image in the mirror.
02

Determine Moth's Distance from the Mirror

The moth is positioned 10 cm in front of the mirror. In a plane mirror, the image distance behind the mirror is equal to the object distance in front of the mirror. So, the moth's image will also be 10 cm behind the mirror.
03

Calculate Total Distance from Your Eyes to Moth's Image

You are 30 cm away from the mirror. To find the apparent distance from your eyes to the moth's image, add the distance from you to the mirror (30 cm) and the distance from the mirror to the moth's image (10 cm).
04

Sum the Distances

Add the two distances: 30 cm (your distance to the mirror) + 10 cm (distance from mirror to moth's image) = 40 cm.
05

Final Answer

The apparent distance between your eyes and the moth's image is 40 cm.

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Key Concepts

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

mirror image distance
In a plane mirror, the distance between the mirror and the image of an object is the same as the distance between the mirror and the object itself. This is known as the 'mirror image distance'. When an object, such as a moth, is placed in front of a plane mirror at a certain distance, its image appears the same distance behind the mirror. So, if the moth is 10 cm in front of the mirror, the image forms 10 cm behind the mirror.

Always visualize a plane mirror reflecting objects as if it's a window through which the object appears at the same distance from your perspective.

This principle is key in many optical devices and setups, helping us understand how we perceive reflected images. By knowing the image distance, we can easily determine the position of an image, making this fundamental concept crucial in mastering reflection problems.
object distance in mirror
To determine the apparent position of an image in a mirror, it's essential to understand the 'object distance in mirror' or simply the distance of the object from the mirror.

The object distance is always measured from the object to the mirror surface. In our exercise, the moth is the object and it is 10 cm away from the plane mirror.

This distance plays a critical part in reflecting the image. The mirror creates an image that seems to exist on the other side of the mirror at the same distance as the object. This behavior doesn't depend on whether the object is in front or behind you – the concept remains consistent.

Knowing the object distance allows you to calculate the total distance from an observer (in this case, you) to the reflected image by simply combining your distance to the mirror and the object's distance to the mirror.
reflection geometry
Reflection geometry deals with how rays of light bounce off reflective surfaces, which is primarily governed by the 'law of reflection'. This law states that the angle of incidence (the angle at which a light ray hits the surface) is equal to the angle of reflection (the angle at which the light ray bounces off).

In our exercise, understanding reflection geometry helps comprehend how the image of the moth forms behind the mirror and at what apparent distance it will appear to you. For plane mirrors, the image seems to be directly behind the mirror at an equal distance to the actual object distance.

This fundamental idea is also used to explain how images are perceived in mirrors in terms of height and orientation – for example, why texts appear reversed in a plane mirror.

Essentially, reflection geometry is the backbone of many principles in optics, defining how we see and interact with the world through reflective surfaces.

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Most popular questions from this chapter

Show that the distance between an object and its real image formed by a thin converging lens is always greater than or equal to four times the focal length of the lens.

You look through a camera toward an image of a hummingbird in a plane mirror. The camera is \(4.30 \mathrm{~m}\) in front of the mirror. The bird is at camera level, \(5.00 \mathrm{~m}\) to your right and \(3.30 \mathrm{~m}\) from the mirror. What is the distance between the camera and the apparent position of the bird's image in the mirror?

A luminous object and a screen are a fixed distance \(D\) apart. (a) Show that a converging lens of focal length \(f\), placed between object and screen, will form a real image on the screen for two lens positions that are separated by a distance $$d=\sqrt{D(D-4 f)}$$ (b) Show that $$\left(\frac{D-d}{D+d}\right)^{2}$$ gives the ratio of the two image sizes for these two positions of the lens.

Address each part of this question in two ways: (1) by drawing and interpreting appropriate geometrical diagrams and (2) by appealing to the lens equation and the expression for lateral magnification and demonstrating your result mathematically. If your two approaches do not agree, explain which one is correct and why the other is wrong. (a) Suppose you are using a camera and wish to have a larger image of a distant object than you are obtaining with the lens currently in use. Would you change to a lens with a longer or a shorter focal length? Explain your reasoning. (Hint: Note that the object distance is essentially fixed.) (b) Suppose you are using a slide projector and wish to obtain a larger image on the screen. You cannot achieve this by moving the screen farther from the projector because you are already using the entire length of the room. Would you change to a lens with a longer or a shorter focal length than the one you are using? Explain your reasoning. (Hint: Note that the image distance is essentially fixed.)

The formula $$\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$$ is called the Gaussian form of the thin-lens formula. Another form of this formula, the Newtonian form, is obtained by considering the distance \(x\) from the object to the first focal point and the distance \(x^{\prime}\) from the second focal point to the image. Show that $$x x^{\prime}=f^{2}$$ is the Newtonian form of the thin-lens formula.

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