Question

Find the position, nature and size of the image formed by a convex lens of focal length 20cm of an object 4cm high placed at a distance 30cm from it.

Answer 1

f=20cm
u=30cm
h=4cm
v=?,h'=?

1/v-1/u=1/f
1/v=1/20+1/-30
1/v=1/20-1/30
1/v=(3-2)/60
1/v=1/60
v=60cm

-v/u=h'/h
-60/30=h'/4
-2×4=h'
h'=-8
h'=8cm, reverse image

Answer 2

Answer:

• The image is formed at a distance of $20 \mathrm{~cm}$from the convex lens (on its left side).
• The value of magnification is more than 1 , the image will be larger than the object.
• The positive sign for magnification suggests that the image is formed above principal axis.
• Height or size of the image is$8 \mathrm{~cm} .$

Explanation:

Given:

Object distance, $u=-10 \mathrm{~cm}$ (It is to the left of the lens.)

Focal length, $f=+20 \mathrm{~cm}$ (It is a convex lens.)

Putting these values in the lens formula, we get:

$1 / v-1 / u=1 / f(v=$ Image distance)

$1 / v-1 /-10)=1 / 20$

or,$v=-20 \mathrm{~cm}$

Thus, the image is formed at a distance of $20 \mathrm{~cm}$from the convex lens (on its left side).

Only a virtual and erect image is formed on the left side of a convex lens. So, the image formed is virtual and erect.

Now,

Magnification,$m=v / u$

$m=-20 /(-10)=2$

Because the value of magnification is more than 1 , the image will be larger than the object. The positive sign for magnification suggests that the image is formed above principal axis.

Height of the object, $h=+4 \mathrm{~cm}$

magnification $m=h^{\prime} / h ( h=$ height of object)

Putting these values in the above formula, we get:

$2=h^{\prime} / 4\left(h^{\prime}=\right.$ Height of the image)

$h^{\prime}=8 \mathrm{~cm}$

Thus, the height or size of the image is$8 \mathrm{~cm} .$