Question

At what distance should an object be placed from a convex lens of focal length 18cm to obtain an image at 24cm from it on the otherside.What will be the magnification produced in this case?​

Answer 1

Answer:

Focal length

Distance of image

Using lens formula to find the position of u.

According to lens formula,

1/f = 1/v-1/u

subtracting the values

1/18 =1/24 - 1/u

u= -72cm

Therefore the object must be placed at a distance of 72cm from the focal length of the lens .

And the magnification of the image formed will be inverted this could be seen by the negative sign obtained. And this sign obtained. And this sign indicates the position in respect to the coordinate where the focal length of the lens is placed.

Answer 2

Answer:

Given:

$\sf{f=18cm}$

$\sf{v=24cm}$

$\sf{Lens=Convex \:Lens}$

Solution:

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

$\sf{{\dfrac{1}{18}}={\dfrac{1}{24}}-{\dfrac{1}{u}}}$

$\sf{{\dfrac{1}{u}}={\dfrac{1}{24}}-{\dfrac{1}{18}}}$

$\sf{{\dfrac{1}{u}}={\dfrac{3-4}{72}}}$

$\sf{{\dfrac{1}{u}}=-{\dfrac{1}{72}}}$

$\sf{u=-72cm}$

$\bold{Magnification:}$

$\sf{Magnification=}{\dfrac{+24}{-72}}$

$\sf{Magnification=-}{\dfrac{1}{3}}$

Thus, the image formed will be real and inverted, and would be of $\sf{\dfrac{1}{3}}$ times the size of object.

The image formed will be smaller than the height of object.