Question

A2.0 cm tall object is placed perpendicular to the principal axis of a
convex lens of focal length 10 cm. The distance of the object from the
lens is 15 cm. Find the nature, position and size of the image. Also
find its magnification.​

Answer 1

h is the height of the

nage and the belona

Image distance i

Height of the image h'

by a lens. then the

Height of the object h =

for spherica

solution

+ 2.0 cm

* + 10 cm

- 15 cm,

u

focal length /

object-distance

1

le object-distance

by

1109

Since

f

1

1

1

+

or

D

(10.10

1 1

1

+

D (-15) 10

tance should

age at 10 cm

the lens.

1 1

15 10

1 -2 +3

D

30

U= + 30 cm

1

30

same side

Or,

The positive sign of u shows that the image is formed at a distance of

30 cm on the other side of the optical centre. The image is real and

inverted,

or,

h' v

Magnification m=

h

h' = h (v/u)

Height of the image, h' = (2.0) (+30/15) = - 4.0 cm

Magnification m= v/u

+ 30 cm

= -2

-15 cm

The negative signs of m and h' show that the image is inverted and

real. It is formed below the principal axis. Thus, a real, inverted image,

4 cm tall, is formed at a distance of 30 cm on the other side of the

lens. The image is two times enlarged.

or, m

Answer 2

Answer:

Explanation:

h=2cm

f=10cm

u=-15cm

Using Lens Formula

1/f=1/v-1/u

1/10=1/v-(-1/15)

1/v=1/10-1/15

1/v=3-2/30

1/v=1/30

v=30cm  is position of image.

magnification (m)= v/u

m=30/-15

m=-2

also, m=h 2(height of image)/h

h2=-2*2=-4cm

Image is enlarged, real, inverted and 30 cm in opposite direction as that of the object.