Question

An object is placed on the principal axis of convex
lens of focallength 30 cm. If the virtual image
height is double the height of object. then the
object is placed at a distance of from the lens.
A​

Answer 1

Answer:

it is paced between pole and principal focus

means less than 30cm

Answer 2

The object is placed at a distance of 15 cm

Explanation:

Given:

Focal length of convex lens f = 30 cm

Image is virtual and its height is double to that of object

To find out:

Distance of object from lens

Solution:

We know that magnification of lens

$m=\frac{v}{u}=\frac{\text{Height of image}}{\text{Height of object}}$

$\implies \frac{v}{u}=2$

$\implies v=2u$

The sign of both image distance and object distance will be same

From lens formula

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

$\frac{1}{-2u}-\frac{1}{-u}=\frac{1}{30}$

$\implies -\frac{1}{2u}+\frac{1}{u}=\frac{1}{30}$

$\implies \frac{1}{2u}=\frac{1}{30}$

$\implies 2u=30$

$\implies u=15$ cm

Hope this answer is helpful.

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