Question

The image of a candle flame formed by a lens is obtained on a
screen placed on the other side of the lens. If the image is three
times the size of the flame and the distance between lens and
image is 80 cm, at what distance should the candle be placed
from the lens? What is the nature of the image at a distance of
80 cm and the lens?

Answer 1

Answer:

As the image is enlarged it is a convex lens

Given:

Magnification=3

v=+/-80cm ( image distance in convex lens can be positive or negative)

u=?

Magnification=v/u

3=-80/u

u=-80/3 or -26.666.... ( the value of v is taken negative because object distance is always negative)

Distance the candle be placed is-26.666.. cm

THE NATURE OF IMAGE :

Virtual

Erect

Enlarged size

CONVEX LENS

hope that this may help u

Answer 2

As the image is obtained on the screen, it is real.

so,

→Magnification , m = –3 ,

→v = 80 cm

→u = ?

As, $\large\bf\pink{m = v/u}$

so,

–3 = 80/u,

u = –80/3 cm .

From

1/f = 1/v – 1/u

=1/80 + 3/80

= 4/80

= 1/20

1/f = 1/20cm

so,

f = 20 cm .

The lens is convex and the image formed at 80 cm from the lens is real and inverted.