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Question
A pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.
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Solution
Given,
Length of the pin = 2.0 cm
Focal length (f) of the lens = 6 cm
As per the question, the centre of the pin is 11 cm away from the lens.
i.e., the object distance (u) = 10 cm
Since, we have to calculate the image of A and B, Let the image be A' and B'
So, the length of the A'B' = size of the image.
Using lens formula: \[\frac{1}{v_A} - \frac{1}{u_A} = \frac{1}{f}\]
Where vA and uA are the image and object distances from point A.
\[\Rightarrow \frac{1}{v_A} - \frac{1}{- 10} = \frac{1}{6}\]
\[\frac{1}{v_A} = \frac{1}{6} - \frac{1}{10} = \frac{1}{15}\]
\[v_A = 15 \text{ cm }\]
Similarly for point B,
Lens formula: \[\frac{1}{v_B} - \frac{1}{u_B} = \frac{1}{f}\]
Where vA and uA are the image and object distances from point B.
\[\frac{1}{v_B} - \frac{1}{- 12} = \frac{1}{6} \]
\[ \Rightarrow v_B = 12 \text{ cm }\]
Length of image = vA − vB = 15 − 12 = 3 cm.
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