An object is placed 30 cm to the left of a diverging lens whose focal length is of magnitude 20 cm. Which one of the following correctly states the nature and position of the virtual image formed?

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Nature of image: inverted, enlarged; distance form lens: 60 cm to the left

Nature of image: inverted, enlarged; distance form lens: 60 cm to the right

Nature of image: erect, diminished; distance form lens: 12 cm to the right

Nature of image: erect, diminished; distance form lens: 12 cm to the left

answer is B.

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Detailed Solution

When an object is placed between 2 f and f (focal length) of the diverging lens, the image is virtual, erect and diminished as shown in the graph. To compute the distance of the image from the lens, we apply

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{- 20} = \frac{1}{v} - \frac{1}{30} \Rightarrow v = - \frac{(20) (30)}{20 + 30}$ = - 12 cm (to the left of the diverging lens.)

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