A candle is placed at a distance of 12cm from of a concave mirror with a focal length of 8cm. The candle is 5cm tall. A. Where is the image located? B. What is the height of the image?

Mirror Equation Calculation: To find the location of the image, we can use the mirror equation: 1/f = 1/do + 1/di where f is the focal length, do is the object distance, and di is the image distance. Given: f = -8cm (negative because it's a concave mirror), do = 12cm. Let's calculate di. 1/di = 1/f - 1/do 1/di = 1/(-8) - 1/12 1/di = -1/8 - 1/12 1/di = (-1*3 - 1*2) / (8*3) 1/di = (-3 - 2) / 24 1/di = -5/24 di = -24/5 di = -4.8cmCharacteristics of Image: The negative sign for di indicates that the image is formed on the same side as the object, which is a characteristic of concave mirrors when the object is placed between the focal point and the mirror. This means the image is virtual and upright.Magnification Equation Calculation: Now, let's find the height of the image using the magnification equation: m = -di/do = hi/ho where m is the magnification, hi is the image height, and ho is the object height. Given: ho = 5cm, di = -4.8cm, do = 12cm. Let's calculate hi. m = -di/do m = -(-4.8)/12 m = 4.8/12 m = 0.4 Now, hi = m * ho hi = 0.4 * 5 hi = 2cm

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A candle is placed at a distance of
12cm from of a concave mirror with a focal length of
8cm. The candle is
5cm tall.
A. Where is the image located?
B. What is the height of the image?

$1$ . A candle is placed at a distance of $12 \mathrm{~cm}$ from of a concave mirror with a focal length of $8 \mathrm{~cm}$ . The candle is $5 \mathrm{~cm}$ tall. $\newline$ A. Where is the image located? $\newline$ B. What is the height of the image?

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Grade 7

Evaluate two-variable equations: word problems

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1

. A candle is placed at a distance of

12 \mathrm{~cm}

from of a concave mirror with a focal length of

8 \mathrm{~cm}

. The candle is

5 \mathrm{~cm}

tall.

\newline

A. Where is the image located?

\newline

B. What is the height of the image?

Mirror Equation Calculation: To find the location of the image, we can use the mirror equation: $\newline$ $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\newline$ where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. $\newline$ Given: $f = -8\,\text{cm}$ (negative because it's a concave mirror), $d_o = 12\,\text{cm}$ . $\newline$ Let's calculate $d_i$ . $\newline$ $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$ $\newline$ $\frac{1}{d_i} = \frac{1}{(-8)} - \frac{1}{12}$ $\newline$ $\frac{1}{d_i} = -\frac{1}{8} - \frac{1}{12}$ $\newline$ $f$ $0$ $\newline$ $f$ $1$ $\newline$ $f$ $2$ $\newline$ $f$ $3$ $\newline$ $f$ $4$
Characteristics of Image: The negative sign for $di$ indicates that the image is formed on the same side as the object, which is a characteristic of concave mirrors when the object is placed between the focal point and the mirror. This means the image is virtual and upright.
Magnification Equation Calculation: Now, let's find the height of the image using the magnification equation: $\newline$ $m = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$ $\newline$ where $m$ is the magnification, $h_i$ is the image height, and $h_o$ is the object height. $\newline$ Given: $h_o = 5\,\text{cm}$ , $d_i = -4.8\,\text{cm}$ , $d_o = 12\,\text{cm}$ . $\newline$ Let's calculate $h_i$ . $\newline$ $m = -\frac{d_i}{d_o}$ $\newline$ $m = -\frac{-4.8}{12}$ $\newline$ $m$ $0$ $\newline$ $m$ $1$ $\newline$ Now, $m$ $2$ $\newline$ $m$ $3$ $\newline$ $m$ $4$