Question

An object is placed at a distance of 30cm from a concave lens of focal length 15cm. List four characteristic (nature, position,etc)of the image formed by the lens

Answer 1

u=-30cm   f= -15cm 
1/f=1/v-1/u
1/v=1/f+1/u
on substituting  v=-10cm
m=v/u
m=-10/-30
m=1/3=.33
so the magnification is positive so it is a virtual erect image .
m isless  than 1 so the image is  diminished .
and v is negative so the image is formed on the same side of the lens.

Answer 2

Answer:

u= -30cm

f= -15 cm

1/f=1/v-1/u

1/-15=1/v-(1/-30)

1/-15=1/v+1/30

1/-15-1/30=1/v

-1/15-1/30=1/v

-2-1/30=1/v

-3/30=1/v

-10=v

so,-10 is between F&O

m=v/u

-10/-30

+0.33

since the magnification is positive the image is virtual and erect . the image is formed above principal axis .

magnification is less than 1 the image is diminished

POSITION : between F&O

NATURE: virtual & erect

SIZE: diminished

The image is formed on same side of the object.

Explanation:Now since the object is placed on 30 cm . It means the object is placed at 2f because f=15cm .In this case when the object is placed at 2f the image in concave lens will form b/w f and o . its nature will be virtual and erect(NOTE: whenever a concave lens is given it will always form an virtual & erect image because it diverges the ray and to form an real image rays should meet at a point.) size= diminished.