Question

an object is placed at a distance of 20 CM in front of a concave lens of focal length 20 cm find the position of image and the magnification of image

Answer 1

Given :

Distance of object = 20cm

Focal length = 20cm

Type of lens : concave

To Find :

Position of image and magnification.

Solution :

❒ Position of image can be calculated by using lens formula which is given by

• $\underline{\boxed{\bf{\orange{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}}}}$

» u denotes position of object

» v denotes position of image

» f denotes focal length

Focal length of concave lens is taken negative and that for convex lens is taken positive.

By substituting the given values;

➙ 1/v - 1/(-20) = 1/(-20)

➙ 1/v = -1/20 - 1/20

➙ 1/v = -2/20

➙ v = -20/2

➙ v = -10 cm

❒ Magnification of image :

➠ m = v/u

➠ m = -10/(-20)

➠ m = 1/2 = 0.5

Type of image : Virtual, Erect and small

Answer 2

Answer:

Given :-

Distance of object=20cm

focal length =20cm

Lens type=concave lens

To find:-

position and magnification of the image

Solution:-

u=-20cm

f=-20cm

v=?

According to lens formula

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

• Substitute the values

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {1}{v}=\dfrac {1}{-20}=\dfrac {1}{-20}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {1}{v}+\dfrac {1}{-20}=\dfrac{1}{-20}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {1}{v}=-\dfrac {1}{20}-\dfrac {1}{-20}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {1}{v}=\dfrac {-2}{20}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {1}{v}=-\dfrac {1}{10}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow v=-10$

$\\\therefore\sf distance\:of\:the\:image\:is\:10cm.$

Again

$\boxed{\sf Magnification_{(m)}=\dfrac {v}{u}}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow m=\dfrac {-10}{-20}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow m=\dfrac {1}{2}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow m=0.5$

$\\\therefore\sf Image\:is\:virtual,\:erect\;and\:small.$