An object is placed at a distance of `30 cm` from a concave lens of focal length `15 cm`. Find the nature and position of the image.

Answer

virtual ; erect, at `-- 10.0 cm`

Explanation

Here, `u = -30 cm, f = -15 cm, v = ?`
As `(1)/(v) - 1/u = (1)/(f)`,
`(1)/(v) + 1/30 = 1/(-15) or (1)/(v) = - 1/15 - 1/30 = (-2 - 1)/30 = (-3)/30 or v = -30/3 = -10.0 cm`
As v is negative, the image must be virtual and erect.