Question

The focal length of converging lens is 10cm. An object is kept at 30cm from the centre of lens. Where will be the image be formed

Answer 1

Answer:

$v = 15$

Explanation:

Given :

(We will be using sign convention of a convex lens.)

focal length = f = +10 cm

object distance = u = -30 cm

image distance = v = ?

Using lens formula :

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

Inserting the values in the equation,

$\frac{1}{10} = \frac{1}{v} - (\frac{1}{ - 30} )$

$\frac{1}{10} = \frac{1}{v} + \frac{1}{30}$

$\frac{1}{10} - \frac{1}{30} = \frac{1}{v}$

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

$\frac{2}{30} = \frac{1}{v}$

$\frac{30}{2} = v$

$15 = v$

(P.S. - Please mark as Brainliest if you are satisfied with my answer.)

Answer 2

Answer:

v=15 cm

Explanation:

We are given that

Focal length of converging lens=10 cm

Object distance=u=-30 cm

We have to find the image distance .

We know that lens formula

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

Substitute the values then we get

$\frac{1}{10}=\frac{1}{v}+\frac{1}{30}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{30}$

$\frac{1}{v}=\frac{3-1}{30}=\frac{2}{30}=\frac{1}{15}$

$v=15 cm$

Hence,the image will be formed at distance  15 cm from the center of converging lens.