Question

A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.

Answer 1

Given,
Height of the object, (h₀) = 2 cm,   
Height of the image, (h₁) = 1 cm
distance between image and object, ( \[-\] u + v) = 40 cm 

We know that,
Magnification is given by:
\[m = \frac{h_i}{h_0} = \frac{v}{u}, \frac{1}{2} = \frac{v}{- u}\]   
u = −2v

Using lens formula,

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\] 

\[ \Rightarrow \frac{1}{v} - \frac{1}{- 2v} = \frac{1}{f}\] 

\[ \Rightarrow \frac{1}{v} + \frac{1}{2v} = \frac{1}{f}\] 

\[ \Rightarrow f = \frac{2v}{3}\]
As per the question,
= −u + v = 40 cm,
= −(−2v) + v = 40 cm, 
= 3v = 40 cm
\[v = 13 . 3 = \frac{40}{3}  cm,       u = \frac{80}{3}\] 
∴ f = 8.9 cm 

Hence, the required focal length is 8.9 cm and object distance is 26.66 cm.