Question

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image..​

Answer 1

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$• \: \rm\:Object\:distance\:(u)=-10\:cm$

$• \: \rm\:Focal\:length\:(f)=15\:cm$

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$• \: \rm\:Image\:distance\:(v)=?$

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Using mirror formula:-

$\red{\underline{\boxed{\bf{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}$

$\rm\longrightarrow\dfrac{1}{v}+\dfrac{1}{-10}=\dfrac{1}{15}$

$\rm\longrightarrow\dfrac{1}{v}-\dfrac{1}{10}=\dfrac{1}{15}$

$\rm\longrightarrow\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{10}$

$\rm\longrightarrow\dfrac{1}{v}=\dfrac{2+3}{30}$

$\rm\longrightarrow\dfrac{1}{v}=\cancel\dfrac{5}{30}$

$\rm\longrightarrow\dfrac{1}{v}=\dfrac{1}{6}$

$\rm\longrightarrow\:v=6\:cm$

Thus,the position of image is at distance of 6 cm from the convex mirror.and nature of image is virtual and erect.