The image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed in front of the mirror at a distance of 60 cm from its pole. What is the nature of the mirror? Find its focal length. If the height of the flame is 2.4 cm, find the height of its image. State whether the image formed is erect or inverted.
CBSE DELHI - SET 3 2017
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Since the image is formed on the screen, the image is real. A concave lens cannot form a real image. Therefore, the lens is convex. Focal length of the convex lens, f = ? Object distance, u = 30 cm Image distance, v = +60 cm Since$\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\therefore \frac{1}{\text{f}}=\frac{1}{60}+\frac{1}{-30}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{60}+\frac{1}{30}$
$\Rightarrow \frac{1}{\text{f}}=\frac{(1+2)}{60}$
$\Rightarrow \frac{1}{\text{f}}=\frac{3}{60}$
orf = +20 cm
The magnification of convex lens, m=vu$\Rightarrow \text{m}=\frac{60}{(-30)}$
$\Rightarrow \text{m} = (-2)$
Magnification, $m=\frac{h_1}{h_0}$ Where $h_i=$ Height of image $h_0=$ Height of object. $. m =\frac{h_1}{2.4}$ \text{m}=\frac{\text{h}_\text{i}}{2.4}$
$\Rightarrow \text{h}_\text{i}=(-2\times2.4)$
$\Rightarrow \text{h}_\text{i}= (-4.8)$
Here, negative sign indicates that the image formed is inverted. Therefore, height of image of candle flame is 4.8cm.
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