Question

PHYSICS QUESTION


A concave lens forms an erect image of 1/3 rd size of the object which is placed at a distance 30cm in front of the lens.

Find:

1.)position of image

2.)focal length of lens​

Answer 1

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 ,

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

$3v = - 30$

Therefore ,

$v = - 10$

The position of image is 10 cm further from the lens

Now using lens formula ,

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

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

So , focal length is 15 cm

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