An object of height 6 cm. is placed at a distance of 20 cm. in front of concave lens of power (-8) dioptre. Find the size of image.

if power is  -8 the lens is concave and the  focal length of the  lens is 1/8 metre or 12.5 centimetre

using lens formulae we get v=33.33cm

m=v/u = 33.33/-20=-1.66

size of image = 10 cm and inverted

power=i/focal length

focal length=8cm

using formula         1/f=1/v-1/u                where u=distance of object from pole.

                                                                                   v=distance of image from pole.

                                                                                   f=focal length of the mirror

following the sign conventions we get,

1/-8=1/v-1/-20

1/-8=1/v+1/20

2-5/40=1/v

v=-40/3.

following the formula

h`/h=v/u                                                                        where h` is the size of the image

                                                                                        and h is the size  of the object

and following the sign conventions

h`/6=-40/3/-20

h`=4cm=size of image.

use lens equation

1/f=1/v-1/u

u = -20cm

f = -12.5 cm = 100/8 cm

v comes out to be -100/13

magnification ,m = v/ u = I / O

I = size of image , O = size of object

I comes out to be 30 / 13 = 2.31 cm

height of object= 6 cm

distance of object = -20 cm

power=100/f cm

-8=100/f

f=-12.5 cm

1/f=1/v-1/u;

-1/12.5=1/v-(-1/20)

1/v = -1/2.5-1/20

on solving v = -100/13cm

m=v/u=h'/h

h'=ht. of image

h=ht.of object

m=magnification

v = distance of image

u=dist. of object

-100/13*-20=v/6

on solving v=2.31cm ans

rate me if ur satisfied