Question

# A $4.5cm$ needle is placed at $12cm$ away from a convex mirror of a focal length of $15cm$ Give the location of the image and the magnification. Describe what happened as the needle is moved further from the mirror.

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Solution

## Step 1. Given data:Height of the needle $\left({h}_{1}\right)$ $=4.5cm$Object distance $\left(u\right)=-12cm$Focal length of the convex mirror $\left(f\right)=15cm$Let image distance = $v$Step 2. Formula used:The value of $v$ can be obtained using the mirror formula$1}{u}+1}{v}=1}{f}\phantom{\rule{0ex}{0ex}}1}{v}=1}{f}-1}{u}........\left(1\right)$Where $v=$image distance, $u=$object distance, $f=$focal length.Also, magnification $m={h}_{2}}{{h}_{1}}=-v}{u}.........\left(2\right)$Where ${h}_{2}=$size of image, ${h}_{1}=$height of object,Step 3. Calculations:Image distance:Now putting the values in equation $\left(1\right)$$1}{v}=1}{15}-1}{\left(-12\right)}\phantom{\rule{0ex}{0ex}}=\left(4+5\right)}{60}=9}{60}\phantom{\rule{0ex}{0ex}}v=60}{9}=6.7cm\phantom{\rule{0ex}{0ex}}$Hence, the image of the needle is $6.7cm$ away from the mirror. Also, it is on the other side of the mirror.Image size:The image size is given by the magnification formula:image is formed $6.7cm$ behind the convex mirror. it must be virtual and erect.if ${h}_{2}$ is size of image , then$m={h}_{2}}{{h}_{1}}=-v}{u}$$m=-6.7}{\left(-12\right)}=0.558$Now ${h}_{2}}{{h}_{1}}=0.558\phantom{\rule{0ex}{0ex}}{h}_{2}=0.558×4.5\phantom{\rule{0ex}{0ex}}{h}_{2}=2.5cm$Hence magnification $m={h}_{2}}{{h}_{1}}=2.5}{4.5}=0.56$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.Thus, (i) Image distance $\left(v\right)=6.7cm$ away from the mirror. (ii) Image size $\left({h}_{2}\right)=2.5cm$ (iii) 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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