Question

an object of 4 cm height is placed 30 cm away from a convex lens of focus length 20cm. and what will be the height of the image.​

Answer 1

Given, h
o
​
=4cm,u=+30cm,f=+20cm
To find: v,h
i
​
and nature of image
From mirror formula, we have

u
1
​
+
v
1
​
=
f
1
​


∴
v
1
​
=
f
1
​
−
u
1
​
=
uf
u−f
​


v=
uf
u−f
​
=
30cm−(20cm)
30cm×20cm
​

=
+10
600cm
​
=+60cm

Magnification, m=
h
o
​

h
i
​

​
=−
u
v
​


or h
i
​
=−
u
v
​
×h
o
​
=−(−
+30fm
+60cm
​
)×4cm=−8cm

Thus, image is formed at a distance of 60cm in front of the mirror. Height of the image is 8cm. As v is positive, the image is real. Since h
i
​
is negative, it is inverted.

Answer 2

★Given:-

• Height of object = 4cm
• Object distance = 30cm
• Focus length = 20cm

★To find:-

• Height of image.

★Solution:-

Using the lens formula,

✦1/v - 1/u = 1/f  

Where we have:

• u = -30cm
• f = 20cm  
• v is unknown

Putting values in the formula,

→1/v - 1/u = 1/f

→1/v - 1/(-30) = 1/20

→1/v + 1/30 = 1/20

→1/v = 1/20 - 1/30

→1/v = 3 - 2 /60

→1/v = 1/60

→v = 60cm

Now,we know:

✦Magnification = Height of image / height of object = v/u

Putting values,

→height of image/4  = 60/-30

→height of image = -2(4)

→height of image = -8cm.

Therefore,

The height of image is -8m.

_____________