Question

what is the focal length of a lens which produces an image three times as large as the object if the distance between the object and the image is 1 m

plz explain with steps
plz

Answer 1

Consider variables u = object
distance, v = image distance
Given,

-u+v=100....eq (1)
Magnification = v/u= -3
v=-3.....eq (2)
Focal length 1/f = 1/v-1/u.........eq (3)


Therefore, f=3u/4 = -3* -25/4 = 18.75cm.

Answer 2

Concept:

The focal length of an optical system is a measure of the distance between the focal point of the lens and the lens itself.

Given:

The image is three times as large as the object.

The distance between the object and the image is $1m$.

Find:

The focal length of the lens.

Solution:

Let $u,v$ be the object distance and the image distance.

Now,

The distance between the object and image:

$-u+v=100$ ( in $cm$)

We know,

$Magnification=\frac{v}{u}=-3$

$v=-3u$

Therefore,

$-u-3u=100\\-4u=100\\u=-25cm$

and, $v=75cm$

Now, the focal length:

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

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

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

$f=\frac{(-25)(75)}{-100}$

$f=18.75cm$

The focal length of the lens is $18.75cm$.

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