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Chapter 25: Problem 22

A small statue has a height of \(3.5 \mathrm{cm}\) and is placed in front of a concave mirror. The image of the statue is inverted, \(1.5 \mathrm{cm}\) tall, and located \(13 \mathrm{cm}\) in front of the mirror. Find the focal length of the mirror.

Short Answer

Expert verified
The focal length of the mirror is approximately \(-22.73 \text{ cm}\).

Step by step solution

01

Understanding the Problem

We have a concave mirror with a real, inverted image. The object's height is given as \(3.5 \text{ cm}\) and the image height is \(1.5 \text{ cm}\), both measured to the vertical axis of the mirror. The image distance is \(-13 \text{ cm}\) (negative because it's a real, inverted image). We need to find the focal length of the mirror, \(f\).
02

Using the Magnification Formula

Start by using the magnification formula: \(m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\). Here, \(h_i = 1.5 \text{ cm}\), \(h_o = 3.5 \text{ cm}\), and \(d_i = -13 \text{ cm}\). Substitute these into the equation: \(\frac{1.5}{3.5} = \frac{-(-13)}{d_o}\).
03

Solving for Object Distance

Simplify the magnification formula to find the object distance \(d_o\): \(\frac{1.5}{3.5} = \frac{13}{d_o}\). Cross-multiply to solve for \(d_o\): \(1.5 \cdot d_o = 3.5 \cdot 13\). Therefore, \(d_o = \frac{3.5 \times 13}{1.5} = 30.33 \text{ cm}\).
04

Applying the Mirror Equation

Use the mirror equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) to find the focal length. Substitute \(d_o = 30.33 \text{ cm}\) and \(d_i = -13 \text{ cm}\) into the equation: \(\frac{1}{f} = \frac{1}{30.33} + \frac{1}{-13}\).
05

Solving the Mirror Equation

Calculate \(\frac{1}{f}\) by performing the addition: \(\frac{1}{30.33} + \frac{1}{-13} = \frac{1}{30.33} - \frac{1}{13}\). This simplifies to \(\frac{1}{f} \approx 0.033 - 0.077 = -0.044\). Therefore, \(f \approx \frac{1}{-0.044} \approx -22.73 \text{ cm}\), indicating the focal length is \(-22.73 \text{ cm}\).

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Key Concepts

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

magnification formula
Understanding the magnification formula is key when working with mirrors. It helps us relate the size of an image to the size of the object it's reflecting. In simple terms, the formula is: \(m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\). Here, \(m\) stands for magnification, \(h_i\) is the image height, \(h_o\) is the object height, \(d_i\) is the image distance, and \(d_o\) is the object distance.
Using the magnification formula, we can find out how much bigger or smaller the image is compared to the object. A negative magnification value indicates the image is inverted, which is common in concave mirrors.
Given the heights \(h_i = 1.5\) cm and \(h_o = 3.5\) cm, along with \(d_i = -13\) cm, this formula helps us figure out the object distance by setting up the equation \( \frac{1.5}{3.5} = \frac{-(-13)}{d_o}\). Solving for \(d_o\) allows us to find where the object is relative to the mirror.
image distance
In mirror problems, especially with concave mirrors, understanding image distance is crucial for finding other properties like object distance and focal length. The image distance \(d_i\) is the distance from the mirror to the image formed. It can be real or virtual, which affects its sign; real images have negative distances.
In the scenario given, the image distance is \(-13\) cm. The negative value here indicates that the image is a real inverted image, seen in front of the mirror. Knowing \(d_i\) allows us to work backward using formulas like the magnification formula to solve for the object distance.
This step precedes using the mirror equation to finally find the focal length, which defines the bending power of the mirror.
mirror equation
The mirror equation is a classic formula used to relate object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) of a mirror: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \).
This formula helps us find the mirror's focal length once we know the object and image distances. It is very useful in practical situations involving mirrors since it connects directly to how mirrors focus light.
In our example, once the object distance \(d_o = 30.33\) cm is found using the magnification formula, the mirror equation is applied: \( \frac{1}{f} = \frac{1}{30.33} + \frac{1}{-13} \).
Simplifying these terms gives us the focal length鈥檚 reciprocal, and reversing this gives the focal length as \(f = -22.73\) cm. The negative sign signifies that the focal length is on the same side as the incoming light, characteristic of concave mirrors.

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Most popular questions from this chapter

Suppose that you are walking perpendicularly with a velocity of \(+0.90 \mathrm{m} / \mathrm{s}\) toward a stationary plane mirror. What is the velocity of your image relative to you? The direction in which you walk is the positive direction.

The image produced by a concave mirror is located \(26 \mathrm{cm}\) in front of the mirror. The focal length of the mirror is \(12 \mathrm{cm} .\) How far in front of the mirror is the object located?

When an object is located very far away from a convex mirror, the image of the object is \(18 \mathrm{cm}\) behind the mirror. Using a ray diagram drawn to scale, determine where the image is located when the object is placed \(9.0 \mathrm{cm}\) in front of the mirror. Note that the mirror must be drawn to scale also. In your drawing, assume that the height of the object is \(3.0 \mathrm{cm}\).

The radius of curvature of a mirror is \(24 \mathrm{cm} .\) A diamond ring is placed in front of this mirror. The image is twice the size of the ring. Concepts: (i) Is the mirror concave or convex, or is either type possible? (ii) How many places are there in front of a concave mirror where the ring can be placed and produce an image twice the size of the object? (iii) What are the possible values for the magnification of the image of the ring? Calculation: Find the object distance of the ring.

The image behind a convex mirror (radius of curvature \(=68 \mathrm{cm}\) ) is located \(22 \mathrm{cm}\) 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.
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