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.

#### 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* **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.