Question

a tree is at a distance of 24 cm from a convex lens. if the focal length of the lens is 18 cm find how far the image will ne found.also state the position and nature of the image. give reason to support your answer​

Answer 1

Given:

$\sf{u=-24cm}$

$\sf{f=18cm}$

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

Solution:

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

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

$\sf{{\dfrac{1}{v}}=-{\dfrac{1}{24}}+{\dfrac{1}{18}}}$

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

$\sf{{\dfrac{1}{v}}={\dfrac{1}{72}}}$

$\sf{v=+72cm}$

$\bold{Magnification:}$

$\sf{Magnification=}{\dfrac{v}{u}}$

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

$\sf{Magnification=-3}$

The positive sign of v shows that the image formed at a distance of 72cm on the other side of optical centre . The image is real & inverted.

The image formed is 3 times the height of object.

Answer 2

The term sublimation is the passage or the transformation or conversion that substances undergo when passing from one state to another, for example from a solid substance to gas. We can define sublimation as the transition of a substance from the solid phase to the gaseous phase without changing into the liquid phase.