Question

a convex lens of focal length 20cm, is used to form an erect image which is twice as large as the object, the position of the object from lens will be​

Answer 1

Answer:

The position of the object from the lens will be 10 cm.

Explanation:

We are given that focal length of convex lens=20 cm

Focal length of convex lens is positive.

We are given that image formed by convex lens is erect  and twice as large as the object.

We have to find the position of object from the lens

It means we have to distance of object from the lens.

Let h be the height of object and height of image =2h

We know that

$\frac{h'}{h}=\frac{v}{u}$

Substitute the value then we get

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

$v=2u$

We know that lens formula

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

Substitute the values then we get

$\frac{1}{2u}-\frac{1}{u}=\frac{1}{20}$

$\frac{1-2}{2u}=\frac{1}{20}$

$\frac{-1}{2u}=\frac{1}{20}$

$u=-10 cm$

Hence, the object is placed  at 10 cm from the lens.

Answer 2

Answer:

The position of the object from the lens will be 10 cm.

Explanation:

We are given that focal length of convex lens=20 cm

Focal length of convex lens is positive.

We are given that image formed by convex lens is erect  and twice as large as the object.

We have to find the position of object from the lens

It means we have to distance of object from the lens.

Let h be the height of object and height of image =2h

We know that

\frac{h'}{h}=\frac{v}{u}hh′=uv

Substitute the value then we get

\frac{2h}{h}=\frac{v}{u}h2h=uv

v=2uv=2u

We know that lens formula

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}v1−u1=f1

Substitute the values then we get

\frac{1}{2u}-\frac{1}{u}=\frac{1}{20}2u1−u1=201

\frac{1-2}{2u}=\frac{1}{20}2u1−2=201

\frac{-1}{2u}=\frac{1}{20}2u−1=201

u=-10 cmu=−10cm

Hence, the object is placed  at 10 cm from the lens.