Question

a concave lens has focal lenght of 15 cm. the image due to the object placed is formed at 10 cm from the lens. the magnification produced by lens is 

Answer 1

focal length, f = -15 cm
image distance , v = -10 cm
Let object distance is u

use lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{-15}=\frac{1}{-10}-\frac{1}{u}$

$\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}$

$\frac{1}{u}=\frac{1}{15}-\frac{1}{10}$

u = -30 cm

thus object distance is 30cm

now, magnification , m = v/u

= (-10)/(-30)

= +0.33

The positive sign shows that image is erect and virtual. the image is one - third of the size of object.

Answer 2

Magnification= -v/-u

Here v=15cm

U= 10cm

Therefore, magnification= +1.5

Optical magnification is the ratio between the apparent size of an object (or its size in an image) and its true size, and thus it is a dimensionless number.