Question

4. An object of 3 cm high is placed at focus of concave lens. Focal length of lens is 30 cm
Calculate position and size of image.​

Answer 1

Solution :-

According to sign convention

Focal length (f) = - 30 cm

Height of the object ( ho ) = 3 cm

Object distance (u) = - 30 cm

Since object is placed at focus

We know that

1/f = 1/v - 1/u

⇒ 1/(-30) = 1/v - 1/(-30)

⇒ - 1/30 = 1/v - (-1/30)

⇒ - 1/30 = 1/v + 1/30

⇒ - 1/30 - 1/30 = 1/v

⇒ - 2/30 = 1/v

⇒ - 1/15 = 1/v

⇒ - 15 = v

⇒ v = - 15 cm

So image is formed at 15 cm from the optic centre on the same side object is placed. Image formed is virtual.

We know that

Magnification m = hi/ho = v/u

⇒ hi/3 = (-15)/(-30)

⇒ hi/3 = 15/30

⇒ hi = (15 * 3)/30

⇒ hi = 15/10

⇒ hi = 1.5 cm

So the image is erect and diminished.

Answer 2

Formulas used:

$= > \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$= > m = \dfrac{v}{u} = \dfrac{hi}{ho}$

Extra information:

For a biconcave lens ,

sign conventions are as follows

f=-f

v=-v

u=-u

unknown value=+ve

For a biconvex lens,

sign conventions are as follows

f=+f

u=-u

v=+v

unknown value=+ve

The same can be followed for mirrors also