Question

A concave lens of focal length 20 cm forms an image of an object kept at a distance of

60 cm from the lens. Find the position, nature and relative size of the image formed by

it. ​

Answer 1

Explanation:

hope it helps u

stay blessed:-)

Answer 2

$\rm{\underline{\underline\color{darkblue}{Given:-}}}$

• Focal length ( f ) = 20cm.

• Object distance ( u ) = - 60 cm.

• Image distance ( v ) = ?

$\rm{\underline{\underline\color{darkblue}{Find:-}}}$

• Find the position, nature and relative size of the image formed .

$\rm{\underline{\underline\color{darkblue}{Solution:-}}}$

We know that,

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

$\bf\dashrightarrow{ \frac{1}{20} = \frac{1}{v} - \frac{1}{( - 60)} }$

$\bf\dashrightarrow{ \frac{1}{20} = \frac{1}{v} + \frac{1}{60} }$

$\bf\dashrightarrow{ \frac{1}{v} = \frac{3 - 1}{60} = \frac{2}{60} = \frac{1}{30} }$

$\bf\dashrightarrow \fbox \pink{v = 30cm}$

Because the object is placed beyond C.

$\rule{258mm}{1mm}$

Learn more in refractive of light :

$\begin{array}{|c|c|} \sf {}\bf Object & \bf {Refractive\:Index} \\\dfrac{\qquad\qquad\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad\qquad}{} \\\sf Air & \sf 1.0003 \\ &\\\sf Ice & 1.31 \\ & \\ \sf Water & 1.33 \\&\\ \sf Alcohol & 1.36 \\&\\ \sf Kerosene & 1.44 \\&\\ \sf Liquefied\:  Quartz & 1.46 \\&\\ \sf Torpene-Tyne\:oil & 1.47 \\&\\ \sf Benzene & 1.50 \\&\\ \sf Crown\:Glass & 1.52 \\&\\ \sf Canada\:Balsum & 1.53 \\&\\ \sf Rock\:Salt & 1.54 \\&\\ \sf Carbon\:Di sulphate  & 1.63 \\&\\ \sf Dark\:Flint\:Glass & 1.65 \\&\\ \sf Ruby & 1.71 \\&\\ \sf Diamond & 2.42    \\ \sf {}\end{array}$

@Shivam