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.

Answer 1

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 v_A and u_A 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 v_A and u_A 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 = v_A − v_B = 15 − 12 = 3 cm.