# An Object Is Placed At A Distance Of 50 Cm From A Concave Lens Of Focal Length 20 Cm. Find The Nature And Position Of The Image.

The focal length of the concave lens is negative.

f = −20 cm

u = −50 cm

We know that :$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$$ $$\Rightarrow \frac{1}{-20} = \frac{1}{v} – \frac{1}{-50}$$

v = -14.3 cm

The nature of the image is virtual and the position of the image is erect.

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