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.

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Solution

Focal length of concave lens is negative.

$f = - 20 c m$

$u = - 50 c m$

We know that :

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{- 20} = \frac{1}{v} - \frac{1}{- 50}$

$v = - 14.3 c m$ .

The image is virtual, erect.

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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.

Focal length of concave lens is negative. f=−20cmu=−50cmWe know that :1f=1v−1u1−20=1v−1−50v=−14.3cm.The image is virtual, erect.

Focal length of concave lens is negative.

$f = - 20 c m$

$u = - 50 c m$

We know that :

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{- 20} = \frac{1}{v} - \frac{1}{- 50}$

$v = - 14.3 c m$ .

The image is virtual, erect.