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Question
A student places a 8.0 cm tall object perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. He obtains a sharp image of the object on a screen placed on the other side of the lens. What will be the nature (inverted, erect, magnified, diminished) of the image he obtains on a screen? Draw ray diagram to justify your answer.
Solution
Focal length of the lens, f = 20 cm
Object distance, u = −30 cm
According to the lens formula,
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30}\]
\[ \Rightarrow v = 60 cm\]
\[\text{ Magnification }, m = \frac{v}{u}\]
\[ \Rightarrow m = \frac{60}{\left( - 30 \right)} = - 2\]
Hence, the image formed is real, inverted and magnified.
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