The diverging lens in part (a) is replaced by a converging lens also of focal length 100mm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverging lens in part (a).
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u = -150mm f = 100mm$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-150}=\frac{1}{100}$
$\frac{1}{\text{v}}=\frac{1}{100}-\frac{1}{150}$
$\frac{1}{\text{v}}=\frac{1}{300}$
$\text{v}=+300\text{mm}$
The image formed by converging lens is real, inverted and magnified (2 times). It is formed behind the converging lens. On the other hand, the image formed by diverging lens is virtual, crect and diminished. It is formed in front of the diverging lens.
$\frac{1}{\text{v}}-\frac{1}{-150}=\frac{1}{100}$
$\frac{1}{\text{v}}=\frac{1}{100}-\frac{1}{150}$
$\frac{1}{\text{v}}=\frac{1}{300}$
$\text{v}=+300\text{mm}$
The image formed by converging lens is real, inverted and magnified (2 times). It is formed behind the converging lens. On the other hand, the image formed by diverging lens is virtual, crect and diminished. It is formed in front of the diverging lens.
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