Question

A short linear object of 2 cm. lies on the axis of a concave mirror of 15 cm. focal length at a distance of 30 cm. from the mirror. What is the size of the image?

Answer 1

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{1mm}\put(1,1.2){\line(1,0){6.8}}\end{picture}$

$\blacksquare\:\:\footnotesize{\underline{\red{Figure}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.4mm}\put(0,0){\line(1,0){42}}\qbezier(30,10)(38,0)(30,-10)\put(22.5,0){\circle*{0.8}}\put(12.5,0){\circle*{0.8}}\linethickness{0.1mm}\put(12.5,0){\vector(0,1){6}}\put(12.5,0){\vector(0,-1){6}}\put(12.5,6){\line(1,0){20}}\put(32.5,-6){\vector(-1,0){10}}\put(12.5,6){\vector(1,0){10}}\put(12.5,-6){\line(1,0){20}}\put(32.5,6){\line(-5,-3){20}}\put(12.5,6){\line(5,-3){20}}\put(32.5,-1.4){o}\footnotesize{\put(13,3){h}}\footnotesize{\put(13,-3){ h' }}\footnotesize{\put(22,-2.1){F}}\footnotesize{\put(10.5,-2.1){C}}\end{picture}$

• $\footnotesize{\text{OF=focus length=f (-ve)}}$

• $\footnotesize{\text{OC=radius of curvature=r (-ve)}}$

• $\footnotesize{\text{but here , OC=object distance=u (-ve)}}$

• $\footnotesize{\text{object height=h (+ve)}}$

• $\footnotesize{\text{image height=h' (-ve)}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{1mm}\put(1,1.2){\line(1,0){6.8}}\end{picture}$

$\blacksquare\:\:\footnotesize{\underline{\red{\text{From the mirror formula }}}}$

$\footnotesize{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}$

$\blacksquare\:\:\footnotesize{\underline{\red{\text{given that : }}}}$

• $\footnotesize{\text{u = -30 cm}}$

• $\footnotesize{\text{f = -15 cm}}$

$\therefore\:\:\footnotesize{\dfrac{1}{v}+\dfrac{1}{-30}=\dfrac{1}{-15}}$

$\implies\therefore\:\:\footnotesize{\dfrac{1}{v}=\dfrac{1}{30}-\dfrac{1}{15}}$

$\implies\therefore\:\:\footnotesize{\dfrac{1}{v}=\dfrac{-1}{30}}$

$\implies\therefore\:\:\boxed{\footnotesize{v=-30}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{1mm}\put(1,1.2){\line(1,0){6.8}}\end{picture}$

$\blacksquare\:\:\footnotesize{\underline{\red{\text{From the magnification formula }}}}$

$\footnotesize{m=\dfrac{-v}{u}=\dfrac{h'}{h}}$

$\blacksquare\:\:\footnotesize{\underline{\red{\text{given that : }}}}$

• $\footnotesize{\text{u = -30 cm}}$

• $\footnotesize{\text{v = -30 cm}}$

• $\footnotesize{\text{f = -15 cm}}$

• $\footnotesize{\text{h = 2 cm}}$

$\therefore\:\:\footnotesize{\dfrac{-(-30)}{-30}=\dfrac{h'}{-2}}$

$\therefore\:\:\boxed{\footnotesize{h'=-2}}$

$\therefore\:\:\footnotesize{\text{The height of the image will be same as object}}$

$\footnotesize{\text{ height i.e., 2 cm. And the -ve sign of h' indicates }}$

$\footnotesize{\text{that the image will be inverted .}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{1mm}\put(1,1.2){\line(1,0){6.8}}\end{picture}$

Answer 2

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