An object of height `2 cm` is placed at a distance of 15 cm in front of a concave lens of power `-10` dioptre. Find the size of the image.

Answer

`0.8 cm`

Explanation

Here, `h_(1) = 2 cm, u = -15 cm, P = -10 D, h_(2) = ?`
Now, `f = 100/P = 100/(-10) = -10 cm`
As `(1)/(v) - 1/u = (1)/(f), (1)/(v) + 1/15 = 1/(-10) or (1)/(v) = -1/10 - 1/15 = (-3 -2)/30 = (-5)/30`
` v = -6 cm`. As v is negative, image is virtual.
As `m = (h_(2))/(h_(1)) = v/u, (h_(2))/2 = (-6)/(-15) = 0.4`
`h_(2) = 0.8 cm`. As `h_(2)` is positive, image is erect.