Question

A lens produces a magnification +0. 33
is converging or diverging lens?
If the focal length is 15cm
Calculate the position of
object?.​

Answer 1

Answer:

it is converging lens

right to pole

Answer 2

The image is located at a distance of 10.06 cm.

Explanation:

The magnification is positive in concave lens when image is virtual and erect.

Focal length of concave lens, f = -15 cm

Magnification of lens is given by :

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

$0.33=\dfrac{v}{u}$

$v=0.33u$...........(1)

Using lens formula we have :

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

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

$\dfrac{1}{0.33u}=\dfrac{1}{u}+\dfrac{1}{(-15)}$

u = -30.5 cm

Put in equation (1) as :

$v=0.33\times (-30.5)$

v = -10.06

So, the image is located at a distance of 10.06 cm.

Learn more,

Lens formula

https://brainly.in/question/2948925