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Question
Solve the following example.
5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
Solution
Given:
Height of object, ho = 5 cm
Object distance, u = -25 cm
Since the lens is converging, thus it is a convex lens.
Focal length of the lens, f = 10 cm
Using lens formula,
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{- 25} = \frac{3}{50}\]
\[ \Rightarrow v = \frac{50}{3} = 16 . 7 cm\]
Thus, the image is formed
16 . 7 cm right of the lens.
Now, we know
\[\frac{v}{u} = \frac{h_i}{h_o}\]
\[ \Rightarrow h_i = \frac{50}{3 \times - 25} \times 5 = -\frac{10}{3} = - 3 . 3 \text{cm}\]
Thus, the size of the image is 3.3 cm. Negative sign shows that the image formed is real and inverted. Hence, the image formed is real and inverted and diminished.
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