Question

a concave lens forms An erect image of 1/3 size of the object which is placed at a distance 30 cm in front of a lens,find position of image, focal length of the lens

Answer 1

Answer: The focal length is 15 cm and the image is formed at 10 cm from the lens.

Explanation:

Given that,

Distance of the object u = -30 cm

Magnification $m =\dfrac{h'}{h}= \dfrac{1}{3}$

We know that,

The magnification is

$m = \dfrac{h'}{h}=\dfrac{v}{u}$

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

$v = -10 cm$

The image is formed at 10 cm from the lens.

Using lens's formula

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

$\dfrac{1}{f}=\dfrac{1}{-15}$

$f = -15 cm$

Hence, The focal length is 15 cm and the image is formed at 10 cm from the lens.

Answer 2

Answer:

Distance of the object from lens = 30 cm

Take this as -u =30

"Formula for finding the magnification is the ratio between the height of the object to the height of the image"

According to given question ,

1/3=v/-30

3v=-30

Therefore ,

V=-10

The position of image is 10 cm further from the lens

Now using lens formula ,

1/f=1/v-1/u

1/f=1/-15

So , focal length is 15 cm