Question

2. A convex lens placed at a distance of 16 cm from
an object produces an image of magnification
m(|m| > 1). If the object is moved towards the lens
by 8 cm, then again an image of magnification m is
obtained. The numerical value of the focal length of
the lens is.....................​

Answer 1

The focal length of convex lens=12 cm

Explanation:

Object distance =-16 cm

Because object distance is always negative

Magnification =m

Where $\mid m\mid >1$

We know that lens formula

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

Where f= Focal length of lens

v=Image distance of  object from the optical center

u=Distance of object from the optical center.

Magnification,m=$\frac{v}{u}$

Substitute the values then we get

$m=\frac{v}{-16}$

$v=-16m$

Substitute the value in the lens formula

$\frac{1}{f}=\frac{1}{-16m}-\frac{1}{-16}$

$\frac{1}{f}=\frac{1}{-16m}+\frac{1}{16}=\frac{-1+m}{16m}$

$f=\frac{16m}{-1+m}$

When m is positive then

$f=\frac{16m}{-1+m}$

When m is negative

$f=\frac{-16m}{-m-1}=\frac{16m}{1+m}$

When object moves toward the lens then the object distance,u =-(16-8)=-8 cm

Magnification of image=m

Magnification, m=$\frac{v}{-8}$

$v=-8m$

Again, substitute the values in the lens formula

$\frac{1}{f}=\frac{1}{-8m}+\frac{1}{8}=\frac{-1+m}{8m}$

$f=\frac{8m}{-1+m}$

Focal length of lens  does not change with object distance

Therefore,

Case 1:

$\frac{8m}{-1+m}=\frac{16m}{m+1}$

$\frac{m+1}{m-1}=2$

$m+1=2m-2$

$1+2=2m-m=m$

$m=3$

Case 2:

$\frac{8m}{m-1}=\frac{16m}{-1+m}$

$1=2$

It is absurd condition.

Therefore, it is not possible.

Substitute the value of m=3

Then, we get

$f=\frac{8m}{-1+m}=\frac{24}{-1+3}=12$

Hence, the focal length of convex lens =12 cm

# Learn more:

What is convex lens

https://brainly.in/question/1333877

Use of convex lens

https://brainly.in/question/3351460 :Answer by  Vaishnavee 2

Answer 2

Explanation:

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