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:
$\frac{1}{v_{\mathrm{A}}}-\frac{1}{u_{\mathrm{A}}}=\frac{1}{f}$
Where v_A and u_A are the image and object distances from point A.
$\Rightarrow \frac{1}{v_{\mathrm{A}}}-\frac{1}{-10}=\frac{1}{6}$
$\frac{1}{v_{\mathrm{A}}}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$
$v_{\mathrm{A}}=15\mathrm{cm}$

Similarly for point B,
Lens formula:

$\frac{1}{v_{\mathrm{B}}}-\frac{1}{u_{\mathrm{B}}}=\frac{1}{f}$
Where v_A and u_A are the image and object distances from point B.
$$\begin{gathered} \frac{1}{v_{\mathrm{B}}}-\frac{1}{-12}=\frac{1}{6} \\ \Rightarrow v_{\mathrm{B}}=12\mathrm{cm} \end{gathered}$$
Length of image = v_A − v_B = 15 − 12 = 3 cm.