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A concave mirror has a focal length of 12 cm. This mirror forms an image located 36 cm in front of the mirror. What is the magnification of the mirror?

Short Answer

Expert verified
The magnification of the mirror is 4.

Step by step solution

01

Identify the Given Values

We are given the focal length of the concave mirror, \( f = 12 \) cm, and the image distance, \( v = -36 \) cm. We use the negative sign for the image distance because in the convention of mirrors, real images have a negative image distance.
02

Recall the Mirror Formula

The mirror formula relates the object distance (\( u \)), the image distance (\( v \)), and the focal length (\( f \)) and is given by: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \].
03

Rearrange the Mirror Formula

Rearrange the mirror formula to solve for \( \frac{1}{u} \): \[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \].
04

Substitute the Known Values

Substitute \( f = 12 \) cm and \( v = -36 \) cm into the equation: \[ \frac{1}{u} = \frac{1}{12} + \frac{1}{36} \].
05

Calculate \( \frac{1}{u} \)

Find a common denominator and perform the calculation: \[ \frac{1}{u} = \frac{3}{36} + \frac{1}{36} = \frac{4}{36} = \frac{1}{9} \].
06

Solve for Object Distance

Invert the fraction to find the object distance \( u \): \( u = 9 \) cm.
07

Calculate Magnification

The magnification \( m \) is given by the formula \( m = -\frac{v}{u} \). Substitute \( v = -36 \) cm and \( u = 9 \) cm into the equation: \( m = -\frac{-36}{9} = 4 \).
08

Interpret the Magnification

Since the magnification is positive and greater than 1, the image is upright and magnified compared to the object.

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Key Concepts

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

Mirror Formula
A concave mirror is a curved mirror that can focus light to form an image. To understand how it works, we utilize the mirror formula. This formula is a powerful relationship that allows us to find important attributes like the object distance, image distance, and the focal length of the mirror. The mirror formula is expressed as:

\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Where:
  • \( f \) is the focal length of the mirror.
  • \( u \) is the object distance, the distance from the mirror to the object.
  • \( v \) is the image distance, the distance from the mirror to the image.
Using this formula, once you know any two of these three quantities, you can calculate the third. It's especially useful in optics problems to determine how mirrors affect light.
Image Distance
The image distance, noted as \( v \), is a key factor in analyzing mirror systems. It measures how far the formed image is from the mirror itself. In our given problem, this distance was provided as \( -36 \) cm.

The negative sign is crucial. In the realm of concave mirrors, a negative image distance signifies that the image is real and located on the same side as the object.
This reflects an established convention in optics where real images, projected onto a screen, are considered negative. Understanding how to interpret this sign is essential when working through mirror calculations.

Real images formed by concave mirrors can often be captured as they converge light at a distinct point, making it crucial for optical devices like telescopes.
Object Distance
Object distance, described by \( u \), is the length from the mirror to the object being reflected. It's crucial for determining the placement of the object in relation to the mirror. Our previous calculations reveal that \( u \) measures 9 cm in this situation.

To find the object distance, we rearranged our mirror formula:
\[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]
Substituting the known values, we executed some basic arithmetic to derive \( u \).
This calculation is important because knowing the object distance helps us predict where the image will form and its properties. Remember, the object distance is always positive in the context of mirrors as it lies on the reflective side of the mirror.
Magnification
Magnification (\( m \)) describes how much larger or smaller the image is compared to the object. It offers insights into how powerful the mirror is at enlarging or reducing an image. Understanding the formula for magnification is pivotal:
\[ m = -\frac{v}{u} \]
In our exercise, substituting the values of \( v = -36 \) cm and \( u = 9 \) cm yields a magnification of 4.

This positive result implies two things:
  • The image is upright.
  • With a value greater than 1, the image is magnified beyond the original object's size.
This understanding is essential for applications like cameras and projection systems, where image magnification impacts the visual outcome. Thus, magnification helps determine the effectiveness of a mirror in real-world applications.

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Most popular questions from this chapter

The outside mirror on the passenger side of a car is convex and has a focal length of 7.0 m. Relative to this mirror, a truck traveling in the rear has an object distance of 11 m. Find (a) the image distance of the truck and (b) the magnification of the mirror.

The image behind a convex mirror (radius of curvature 68 cm) is located 22 cm from the mirror. (a) Where is the object located and (b) what is the magnification of the mirror? Determine whether the image is (c) upright or inverted and (d) larger or smaller than the object.

The image produced by a concave mirror is located 26 cm in front of the mirror. The focal length of the mirror is 12 cm. How far in front of the mirror is the object located?

In an experiment designed to measure the speed of light, a laser is aimed at a mirror that is 50.0 km due north. A detector is placed 117 m due east of the laser. The mirror is to be aligned so that light from the laser reflects into the detector. (a) When properly aligned, what angle should the normal to the surface of the mirror make with due south? (b) Suppose the mirror is misaligned, so that the actual angle between the normal to the surface and due south is too large by 0.004. By how many meters (due east) will the reflected ray miss the detector?

An object is located 14.0 cm in front of a convex mirror, the image being 7.00 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

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