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प्रश्न
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.
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उत्तर
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.
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