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Light — PHYSICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9PHYSICSLight4 Marks
Question
A converging mirror, forms a three times magnified virtual image when an object is placed at a distance of 8 cm from it. Calculate:
1. the position of the image
2. the focal length of the mirror.
✓

Answer

Let size of the object $= h _{ 0 }= x$
Size of image $=h_1=+3 \times$ (Virtual image)
Distance of the object from the mirror $= u =-8 cm$
Distance of the image from the mirror $= v =$ ?
Focal length of the mirror $=f=$ ?
(i) Magnification $=m=\frac{h_i}{h_0}=-\frac{v}{u}$
$
\begin{array}{l}
\Rightarrow \quad \frac{3 x}{x}=\frac{-v}{-8} \\
v=8 \times 3=24 cm
\end{array}
$
(ii) Using mirror formula :
$
\begin{aligned}
& \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\
& \frac{1}{-8}+\frac{1}{24}=\frac{1}{f} \\
& \frac{1}{f}=-\frac{1}{8}+\frac{1}{24} \\
& \frac{1}{f}=\frac{-3+1}{24}=\frac{-2}{24}=-\frac{1}{12} \\
\Rightarrow & f=-12 cm
\end{aligned}
$

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