Question

Find the position, nature of size of the image formed be a convex lens of focal length 20 cm, of an object 3 cm high placed at a distance of 10 cm from it?

Answer 1

Given :

➳ Focal length = 20cm

➳ Distance of object = 10cm

➳ Height of object = 3cm

➳ Type of lens : $\tt{convex}$

To Find :

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

SoluTioN :

፨ Focal length of convex lens is taken positive.

❍ Position of image :

➝ 1/v - 1/u = 1/f

➝ 1/v - 1/(-10) = 1/20

➝ 1/v = 1/20 - 1/10

➝ 1/v = (2 - 4) / 40

➝ 1/v = -2/40

➝ v = -20cm

❍ Size of image :

⇒ m = -20/-10 = h'/3

⇒ 2 = h'/3

⇒ h' = 6cm

❍ Nature of image :

• Virtual
• Erect
• Enlarged

Answer 2

Here ,

Object distance (u) = -10 cm

Focal length (f) = 20 cm

Object height (h) = 3 cm

We know that , the lens formula is given by

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

Thus ,

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

1/20 - 1/10 = 1/v

(1 - 2)/20 = 1/v

v = -20 cm

∴The position of image is 20 cm from the lens on the left side

Now , the linear magnification of lens is given by

$\boxed{ \sf{Magnification \: (m) = \frac{ h_{i} }{h_{o} } = \frac{ v}{u} }}$

Thus ,

hi/3 = -20/-10

hi = 2 × 3

hi = 6 cm

∴

• The size of image is 6 cm
• The image is virtual and erect

Remmember :

# if m or v is positive , the image is virtual and erect

# if m or v is negative , the image is real and inverted