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# A square of side $3cm$ is located at a distance of $25cm$ from a concave mirror of the focal length of $10cm$. The center of the square is at the axis of the mirror and the plane is normal to the axis of the mirror. The area enclosed by the image of the square is

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## Step 1: Given data Height of the object: ${h}_{\circ }=3cm$Width of the object: ${b}_{o}=3cm$Distance of object from mirror: $u=-25cm$Focal length: $f=-10cm$Let distance of image from mirror be $v$It is given that the rods of the mirror is normal to the square and thus the mirror is parallel to the square.Step 2: Formula used Mirror formula is given as,$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$The magnification produced by a mirror,$m=\frac{{h}_{i}}{{h}_{\circ }}=\frac{-v}{u}$where ${h}_{i}$ is the height of the image and ${h}_{o}$ is the height of the objectStep 3: SolutionSubstituting the values in the mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$\begin{array}{cc}& \frac{1}{v}+\frac{1}{-25}=\frac{1}{-10}\\ ⇒& \frac{1}{v}=\frac{1}{25}-\frac{1}{10}\\ ⇒& v=-\frac{50}{3}cm\end{array}$Therefore magnification is, $m=\frac{{h}_{i}}{{h}_{\circ }}=\frac{-v}{u}$ $\frac{{h}_{i}}{3}=-\left(\frac{-\frac{50}{3}}{-25}\right)\phantom{\rule{0ex}{0ex}}{h}_{i}=-2cm$The negative sign of the height indicates that the image is invertedSimilarly, ${b}_{i}=-2cm$Therefore, the area of the image formed ${A}_{i}={h}_{i}{b}_{i}\phantom{\rule{0ex}{0ex}}{A}_{i}=\left(2\right)×\left(2\right)\phantom{\rule{0ex}{0ex}}{A}_{i}=4c{m}^{2}$Hence, the area enclosed by the image of the square is $4c{m}^{2}$.  Suggest Corrections  0      Similar questions  Explore more