A pin of length 2.0cm lies along the principal axis of a converging lens, the centre being at a distance of 11cm from the lens. The focal length of the lens is 6cm. Find the size of the image.
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Now we have to calculate the image of A and B.
Let the images be A',B'.
So, length of A'B' = size of image.
For A, u = -10cm, f = 6cm
Since, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\frac{1}{-10}=\frac{1}{6}$
⇒ v = 15cm = OA'
For B, u = -12cm, f = 6cm
Again, $\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{12}$
⇒ v = 12cm = OB'
$\therefore$ A'B' = OA' - OB' = 15 - 12 = 3cm.
So, size of image = 3cm.
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