Question

A 2cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10cm. The distance of the object from the lens is 15cm.find the nature, position and size of the image (30cm,-4cm)

Answer 1

Given that, the height of the object (ho) = 2 cm

Focal length (f) = 10 cm

Object distance from lens (u) = -15w cm

We have to find the nature, position and size of the image.

Lens formula:

1/f = 1/v - 1/u

Substitute the known values

→ 1/10 = 1/v - 1/(-15)

→ 1/10 - 1/15 = 1/v

→ 1/v = (3 - 2)/30

→ 1/v = 1/30

→ v = 30 cm

Now,

m = v/u = hi/ho

Substitute the known values

→ 30/(-15) = hi/2

→ -2 = hi/2

→ hi = -2(2)

→ hi = -4 cm

Therefore, the image distance from the lens is 30 cm, the height of the image is -4 cm and the nature of the image is real & inverted.

Answer 2

Given:-

• A 2 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.

To find:-

The nature, position and size of the image.

Solution:-

Height of the object (ho) = 2 cm

Focal length (f) = 10 cm

Object distance (u) = -15 cm

Using lens formula:-

$\boxed{\frac{1}{v} -\frac{1}{u}= \frac{1}{f} }$

$\bold{\frac{1}{v}-\frac{1}{-15} = \frac{1}{10}}$

$\bold{\frac{1}{v}=\frac{1}{10} -\frac{1}{15}}$

$\bold{\frac{1}{v}=\frac{3-2}{30}}$

$\bold{\frac{1}{v}=\frac{1}{30}}$

$\bold{v=30\ cm}$

Now,

Magnification(m) = v/u = hi/ho

⇒ $\bold{\frac{30}{-15}=\frac{hi}{2}}$

⇒ $\bold{-2=\frac{hi}{2} }$

⇒ $\bold{hi=-4\ cm}$

∴ The image distance is 30 cm and the size of the image is -4 cm.

• As the image size is negative, the image is real & inverted.

• The image is magnified as image size > object size.