A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
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Height of the needle, $h_1 = 4.5$ cm
Object distance, $u = – 12$ cm
Focal length of the convex mirror, $f = 15$ cm
Image distance $= v$
The value of v can be obtained using the mirror formula:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$
$=\frac{1}{-25}-\frac{1}{5}=\frac{-5-1}{25}=\frac{-6}{25}$
$\therefore \ \text{u}=-\frac{25}{6}=-4.167 \ \text{cm}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$
$\therefore \ \text{v}=\frac{60}{9}=6.7 \ \text{cm}$
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\therefore \ \text{h}_2=-\frac{\text{v}}{\text{u}}\times\text{h}_1$
$=\frac{-6.7}{-12}\times4.5=+2.5 \ \text{cm}$
Hence, magnification of the image, $\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{2.5}{4.5}=0.56$
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
Object distance, $u = – 12$ cm
Focal length of the convex mirror, $f = 15$ cm
Image distance $= v$
The value of v can be obtained using the mirror formula:
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$
$=\frac{1}{-25}-\frac{1}{5}=\frac{-5-1}{25}=\frac{-6}{25}$
$\therefore \ \text{u}=-\frac{25}{6}=-4.167 \ \text{cm}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$
$\therefore \ \text{v}=\frac{60}{9}=6.7 \ \text{cm}$
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
$\text{m}=\frac{\text{h}_2}{\text{h}_1}=-\frac{\text{v}}{\text{u}}$
$\therefore \ \text{h}_2=-\frac{\text{v}}{\text{u}}\times\text{h}_1$
$=\frac{-6.7}{-12}\times4.5=+2.5 \ \text{cm}$
Hence, magnification of the image, $\text{m}=\frac{\text{h}_2}{\text{h}_1}=\frac{2.5}{4.5}=0.56$
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
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