an object 3cm high is placed prependicularto the principal axis of concave lens of focal length 15cm.the image is form at a distance of 10cm from the lens.find the distance at which object is placed and size and nature of image.?
Focal length (f) = -20 cm
Here,
Object Distance (u) = ?
ho = 3 cm
Image Distance, v = -10 cm
now according to lens formula,
1/f =1/v - 1/u
1/u = 1/v - 1/f
This gives,
u = -30 cm
Magnification,
m = hi/ho= v/u
m = 10/30 = hi/ho
hi =1/3 x 3=1 cm
Image is virtual , erect and smaller.