Question

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

Answer 1

u=-30cm
f=-15
1/v=1/f+1/u=-1/15-1/30=-1/10
v=-10
magnification (m)=v/u=-10/-30=.33
As, m is positive image formed is virtual and erect.
As, m<1 image formed is small and diminished.
As, v is -ve image formed is at the same side of the lens.

Answer 2

Answer:

Given :

Object distance u = 30 cm

Focal length = 15 cm .

Since it's concave lens.

= > u = - 30 and f = - 15 cm

We have lens formula :

1 / f = 1 / v - 1 / u

1 / v = 1 / f + 1 / u

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

1 / v = - 1 / 10

v = - 10 cm .

Now nature of image is as :

( i )  Image is formed on the left side of the lens.

( ii ) The image is virtual and erect.