Question

if the cocave mirror has focal lenth of length of 10 cm what is the position ,nature and magnification of the mirror if the object is placed 20cm from the mirror.​

Answer 1

Given :-

◉$\quad$ Focal length,f = - 10 cm

◉$\quad$ Object distance, u = - 20 cm

To Find :-

◉$\quad$ Magnification, Position, Nature of the mirror.

Solution :-

$\setlength{ \unitlength}{1cm} \thicklines \begin{picture}(10,10) \put(0,0){ \line(1,0){6}} \put(3,0){ \circle*{0.1}} \qbezier(4.5,1.5)(6,0)(4.5, - 1.5) \put(5.25,0){\circle*{0.1}} \put(4.125,0){\circle*{0.1}} \put(3, - 0.4){c} \put(4.1, - 0.4){f} \put(4.5,0.1){ \vector( - 1,0){0.5}}\put(4.4,0.1){ \vector(1,0){0.78}} \put(4.2,0.2){ \footnotesize{10 cm}} \put(3,0){ \vector(0,1){0.8}}\put(4, - 1){ \vector( - 1,0){1}}\put(3.9, - 1){ \vector(1,0){1.2}} \put(3.2, - 1.3){ \footnotesize{u = 20 cm}} \end{picture}$

$\quad$

We know,

$\qquad \boxed{ \sf \frac{1}{f} = \frac{1}{u} + \frac{1}{v} }$

Substituting values, we get

$\sf \Rightarrow \quad - \dfrac{1}{ 10} = - \dfrac{1}{20} + \dfrac{1}{v} \\ \\ \sf \Rightarrow \quad \frac{1}{v} = \frac{1}{20} - \frac{1}{10} \\ \\ \sf \Rightarrow \quad \frac{1}{v} = \frac{1 - 2}{20} \\ \\ \sf \Rightarrow \quad \frac{1}{v} = - \frac{1}{20} \\ \\ \therefore \qquad v = - 20 \sf \: cm$

Also,

$\qquad \boxed{ \sf Magnification = - \dfrac{v}{u}}$

Substituting values, we get

$\sf \rightarrow \quad m = - \bigg( \dfrac{ \cancel{- 20}}{ \cancel{ - 20}} \bigg) \\ \\ \sf \rightarrow \quad m = - 1$

Hence,

• Position of Image : 20 cm from the mirror
• Nature of Image : Real and Inverted
• Magnification : -1
• Size of Image : Same sized as Object

Some Information :-

☛$\quad$ If the object is at the centre of curvature then the image will also be at the centre of curvature and will be of same size, and inverted.

☛$\quad$ The focal length of a concave mirror is always negative and while convex mirror has positive focal length.

☛ $\quad$ The object is always at the left side hence, It will be always negative in both mirrors.